5.1 What a determinant should be

A square matrix $𝐴 \in 𝐹^{𝑛 \times 𝑛}$ gives a linear map $𝑇_{𝐴} : 𝐹^{𝑛} \to 𝐹^{𝑛}$ . Over $ℝ$ we can watch $𝑇_{𝐴}$ act on a region of the plane or of space: it stretches the region, shears it, and possibly turns it over. Two features of this action are recorded by a single number, the determinant $det 𝐴$ . Its absolute value is the factor by which $𝑇_{𝐴}$ scales areas (or volumes), and its sign says whether $𝑇_{𝐴}$ reverses orientation. We will not start from a formula for $det 𝐴$ . Instead, this section lists the properties that a signed area must have, takes them as axioms, and shows that for $2 \times 2$ matrices they leave exactly one possibility, $𝑎 𝑑 - 𝑏 𝑐$ . The same axioms determine $det 𝐴$ for $𝑛 \times 𝑛$ matrices as well, but the proof needs permutations (Section 5.2) and is given in Section 5.3.

5.1.1Signed area

Let $𝐴 \in ℝ^{2 \times 2}$ have columns $𝐚_{1} = 𝐴 𝐞_{1}$ and $𝐚_{2} = 𝐴 𝐞_{2}$ . By linearity, $𝑇_{𝐴} (𝑠 𝐞_{1} + 𝑡 𝐞_{2}) =$ $𝑠 𝐚_{1} +$ $𝑡 𝐚_{2}$ , so $𝑇_{𝐴}$ sends the unit square

𝑄 = {𝑠 𝐞_{1} + 𝑡 𝐞_{2} | 0 \leq 𝑠, 𝑡 \leq 1}

onto the parallelogram

𝑃 (𝐚_{1}, 𝐚_{2}) = {𝑠 𝐚_{1} + 𝑡 𝐚_{2} | 0 \leq 𝑠, 𝑡 \leq 1} .

The same computation shows that $𝑇_{𝐴}$ sends each small grid square $𝐩 + ℎ 𝑄$ onto the parallelogram $𝐴 𝐩 + ℎ 𝑃 (𝐚_{1}, 𝐚_{2})$ . So $𝑇_{𝐴}$ multiplies the area of every grid square by the same factor, the area of $𝑃 (𝐚_{1}, 𝐚_{2})$ , and approximating a region by grid squares suggests that $𝑇_{𝐴}$ multiplies every area by this factor. (Making the approximation precise is a matter for analysis. We use the picture only as motivation; Section 9.2 returns to it.)

Figure 5.1.1. The unit square (dashed) and a letter F drawn in it, under four linear maps; the arrows are the images of

𝐞_{1}

and

𝐞_{2}

. Top row:

𝐼_{2}

, and

𝐴 = (\begin{matrix} 2 & - 1 \\ 1 & 1 \end{matrix})

, which triples areas. Bottom row:

𝑅 = (\begin{matrix} - 1 & 0 \\ 0 & 1 \end{matrix})

, which preserves areas but reverses orientation, and

𝐵 = (\begin{matrix} 1 & 2 \\ 1 & 2 \end{matrix})

, which flattens the square. Blue shading means orientation is preserved, red that it is reversed.

Figure 5.1.1 shows four examples. The map $𝑇_{𝐴}$ triples areas. The reflection $𝑇_{𝑅}$ preserves areas but draws the F backwards, and no rotation of the plane turns a backwards F into an F. The map $𝑇_{𝐵}$ squashes the square onto a segment, so it multiplies areas by $0$ . This matches the fact that $𝐵$ is not invertible: its second column is twice its first, so its columns are linearly dependent.

To tell $𝑇_{𝑅}$ apart from the identity we use orientation. In $𝑄$ , turning from $𝐞_{1}$ to $𝐞_{2}$ is a counterclockwise turn through an angle between $0$ and $𝜋$ . Under $𝑇_{𝐴}$ this is still true of the turn from $𝐚_{1} = (2, 1)$ to $𝐚_{2} = (- 1, 1)$ , but under $𝑇_{𝑅}$ the turn from $𝐚_{1} = (- 1, 0)$ to $𝐚_{2} = (0, 1)$ is clockwise. We say that $𝑇_{𝐴}$ preserves orientation and $𝑇_{𝑅}$ reverses it. Accordingly, the signed area of $𝑃 (𝐚_{1}, 𝐚_{2})$ is its area when the turn from $𝐚_{1}$ to $𝐚_{2}$ is counterclockwise, the negative of its area when the turn is clockwise, and $0$ when $𝐚_{1}$ and $𝐚_{2}$ lie on one line through the origin. The determinant of a real $2 \times 2$ matrix should be the signed area of the parallelogram spanned by its columns. Its absolute value is then the area scaling factor of $𝑇_{𝐴}$ , and its sign tells whether $𝑇_{𝐴}$ preserves orientation.

Intuition (Signed area is linear in each column). Write $𝑆 (𝐚_{1}, 𝐚_{2})$ for the signed area of $𝑃 (𝐚_{1}, 𝐚_{2})$ . We do not yet have a formula for $𝑆$ , but three of its properties can be read off from pictures.

$𝑆 (𝐞_{1}, 𝐞_{2}) = 1$ , since $𝑃 (𝐞_{1}, 𝐞_{2})$ is the unit square.
$𝑆 (𝐯, 𝐯) = 0$ for every $𝐯$ , since $𝑃 (𝐯, 𝐯)$ is a segment.
$𝑆$ is linear in each argument when the other argument is fixed.

The first two are clear. For the third, compute area as base times height. Fix $𝐚_{1} \neq 𝟎$ as the base, and let $ℓ$ be the line through $𝟎$ and $𝐚_{1}$ . The area of $𝑃 (𝐚_{1}, 𝐚_{2})$ is the length of $𝐚_{1}$ times the distance from $𝐚_{2}$ to $ℓ$ . Count that distance as positive when $𝐚_{2}$ lies on the side of $ℓ$ that $𝐚_{1}$ turns toward when it turns counterclockwise, and negative on the other side. With this signed height $ℎ (𝐚_{2})$ , the signed area is the length of $𝐚_{1}$ times $ℎ (𝐚_{2})$ . If $𝐚_{1}$ points along the positive $𝑥$ -axis, then $ℎ (𝑥, 𝑦) = 𝑦$ , a linear function of $(𝑥, 𝑦)$ . Rotating the whole picture changes neither areas nor orientations, so $ℎ$ is linear whatever the direction of $𝐚_{1}$ , and so is $𝑆 (𝐚_{1}, \cdot)$ . Finally, reversing the order of $𝐚_{1}$ and $𝐚_{2}$ reverses the direction of the turn, so $𝑆 (𝐚_{2}, 𝐚_{1}) =$ $- 𝑆 (𝐚_{1}, 𝐚_{2})$ , and linearity in the second argument gives linearity in the first.

Two special cases are worth seeing. Adding a multiple of $𝐚_{1}$ to $𝐚_{2}$ slides $𝐚_{2}$ parallel to $ℓ$ , which changes neither the base nor the height, so $𝑆 (𝐚_{1}, 𝐚_{2} + 𝑐 𝐚_{1}) =$ $𝑆 (𝐚_{1}, 𝐚_{2})$ . Moving $𝐚_{2}$ across $ℓ$ changes the sign of $𝑆$ . Try both below.

Drag

𝐴 𝐞_{2}

, or focus it and use the arrow keys. Sliding it parallel to

𝐴 𝐞_{1}

leaves the signed area unchanged, and moving it across the line through

𝐴 𝐞_{1}

flips the sign; the letter F in the square is mirrored exactly when the sign is negative. The formula for the signed area in terms of the entries comes out of the axioms in Theorem 5.1.13.

The signs are what make linearity possible. Unsigned area is not linear in either argument: $𝑃 (𝐞_{1}, 𝐞_{2})$ and $𝑃 (- 𝐞_{1}, 𝐞_{2})$ both have area $1$ , but $𝑃 (𝐞_{1} + (- 𝐞_{1}), 𝐞_{2}) =$ $𝑃 (𝟎, 𝐞_{2})$ has area $0$ , which is not $1 + 1$ . With signs, $𝑆 (𝐞_{1}, 𝐞_{2}) +$ $𝑆 (- 𝐞_{1}, 𝐞_{2}) =$ $1 +$ $(- 1) =$ $0$ , as linearity requires. So any function that agrees with area up to sign and is linear in each argument must take negative values, and the sign it carries is the orientation.

5.1.2The determinant axioms

The three properties of $𝑆$ make no reference to area or to $ℝ$ , and they make sense for $𝑛$ vectors in $𝐹^{𝑛}$ over any field. We take them as the definition. For a function $𝐷$ on $𝐹^{𝑛 \times 𝑛}$ and a matrix $𝐴$ with columns $𝐚_{1}, \dots, 𝐚_{𝑛}$ , we write $𝐷 (𝐚_{1}, \dots, 𝐚_{𝑛})$ for $𝐷 (𝐴)$ . This lets us treat $𝐷$ as a function of $𝑛$ vectors in $𝐹^{𝑛}$ , the columns of its argument.

Definition 5.1.1 (Determinant function). Let $𝑛 \geq 1$ , and let $𝐷 : 𝐹^{𝑛 \times 𝑛} \to 𝐹$ .

$𝐷$ is normalized if $𝐷 (𝐼_{𝑛}) = 1$ , that is, $𝐷 (𝐞_{1}, \dots, 𝐞_{𝑛}) = 1$ .
$𝐷$ is alternating if $𝐷 (𝐴) = 0$ whenever two columns of $𝐴$ are equal, that is, whenever $𝐚_{𝑖} = 𝐚_{𝑗}$ for some $𝑖 \neq 𝑗$ .
$𝐷$ is multilinear if it is linear in each column when the other columns are held fixed. That is, for each $𝑗 \in [𝑛]$ and all fixed $𝐚_{𝑖}$ with $𝑖 \neq 𝑗$ , the map $𝐹^{𝑛} \to 𝐹$ given by
$𝐱 \mapsto 𝐷 (𝐚_{1}, \dots, 𝐚_{𝑗 - 1}, 𝐱, 𝐚_{𝑗 + 1}, \dots, 𝐚_{𝑛})$
is linear. Concretely,
$𝐷 (\dots, 𝑐 𝐱 + 𝐲, \dots) = 𝑐 𝐷 (\dots, 𝐱, \dots) + 𝐷 (\dots, 𝐲, \dots)$
for all $𝐱, 𝐲 \in 𝐹^{𝑛}$ and $𝑐 \in 𝐹$ , where the dots stand for the same fixed columns in all three places.

A determinant function on $𝐹^{𝑛 \times 𝑛}$ is a function $𝐷 : 𝐹^{𝑛 \times 𝑛} \to 𝐹$ that is normalized, alternating, and multilinear. These three conditions are the determinant axioms.

So a determinant function is what the motivation asks for: the unit cube has volume $1$ , a parallelepiped with two equal edges is flat and has volume $0$ , and signed volume is linear in each edge. In Section 5.3 we prove that for every $𝑛$ there is exactly one determinant function on $𝐹^{𝑛 \times 𝑛}$ (Theorem 5.3.4). It is called the determinant and written $det$ . Until then, $𝐷$ denotes a function satisfying some of the axioms, and each result below says which ones it uses. Several results need only multilinearity and alternation, so they apply to functions such as $𝐷 = 0$ as well.

Example 5.1.2 (

𝑛 = 1

). A

1 \times 1

matrix is a scalar, so

𝐷 : 𝐹 \to 𝐹

. The alternating condition is empty, since there is only one column. Multilinearity says that

𝐷

is linear, so

𝐷 (𝑎) =

𝐷 (𝑎 \cdot 1) =

𝑎 𝐷 (1)

for every

𝑎

, and normalization says

𝐷 (1) = 1

. Hence

𝐷 (𝑎) = 𝑎

is the only determinant function on

𝐹^{1 \times 1}

, and it is one.

Non-example 5.1.3 (One axiom at a time). Write $𝐴 =$ $(\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix}) \in$ $𝐹^{2 \times 2}$ , whose columns are $(𝑎, 𝑐)$ and $(𝑏, 𝑑)$ . Each function below satisfies two of the axioms and fails the third.

(a) $𝐷 (𝐴) = 𝑎 𝑑$ is normalized, and it is multilinear: for a fixed second column, $(𝑎, 𝑐) \mapsto 𝑑 𝑎$ is linear, and for a fixed first column, $(𝑏, 𝑑) \mapsto 𝑎 𝑑$ is linear. It is not alternating, since $𝐷 (\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}) = 1$ .
(b) $𝐷 = 0$ is alternating and multilinear but not normalized.
(c)Over $ℝ$ , the unsigned area $𝐷 (𝐴) =$ $| 𝑎 𝑑 -$ $𝑏 𝑐 |$ is normalized and alternating (if the columns are equal, then $𝑎 𝑑 -$ $𝑏 𝑐 =$ $𝑎 𝑐 -$ $𝑐 𝑎 =$ $0$ ). It is not multilinear: $𝐷 (- 𝐞_{1}, 𝐞_{2}) = 1$ , while linearity in the first column would force $𝐷 (- 𝐞_{1}, 𝐞_{2}) =$ $- 𝐷 (𝐞_{1}, 𝐞_{2}) =$ $- 1$ .

So none of the three axioms follows from the other two.

Remark 5.1.4. Multilinear does not mean linear. A multilinear

𝐷

is linear in one column at a time, and scaling the whole matrix scales every column, so

𝐷 (𝑐 𝐴) =

𝑐^{𝑛} 𝐷 (𝐴)

, not

𝑐 𝐷 (𝐴)

. Similarly

𝐷 (𝐴 + 𝐵)

is usually not

𝐷 (𝐴) + 𝐷 (𝐵)

; Exercise 5.1.21 has an example.

Exercise 5.1.5.routine Let

𝐷 : 𝐹^{𝑛 \times 𝑛} \to 𝐹

be multilinear. Show that

𝐷 (𝑐 𝐴) =

𝑐^{𝑛} 𝐷 (𝐴)

for every

𝑐 \in 𝐹

and

𝐴 \in 𝐹^{𝑛 \times 𝑛}

. In particular

𝐷 (- 𝐴) =

{(- 1)}^{𝑛} 𝐷 (𝐴)

, and

𝐷 (𝐴) = 0

whenever

𝐴

has a zero column.

Solution.

Uses. Definition 5.1.1 (Determinant function), Proposition 2.1.4 (Linear maps preserve $𝟎$ and negatives)

Hint 1. The columns of

𝑐 𝐴

are

𝑐 𝐚_{1}, \dots, 𝑐 𝐚_{𝑛}

. Pull the scalars out one column at a time.

Hint 2. For the zero column, a linear map sends

𝟎

0

. Alternatively,

𝟎 = 0 \cdot 𝟎

Sketch. Linearity in column

𝑗

pulls out one factor

𝑐

, for each of the

𝑛

columns. A zero column

𝟎 = 0 \cdot 𝟎

pulls out a factor

0

The columns of $𝑐 𝐴$ are $𝑐 𝐚_{1}, \dots, 𝑐 𝐚_{𝑛}$ . By linearity in the first column, then the second, and so on,

𝐷 (𝑐 𝐚_{1}, 𝑐 𝐚_{2}, \dots, 𝑐 𝐚_{𝑛}) = 𝑐 𝐷 (𝐚_{1}, 𝑐 𝐚_{2}, \dots, 𝑐 𝐚_{𝑛}) = \dots = 𝑐^{𝑛} 𝐷 (𝐚_{1}, \dots, 𝐚_{𝑛}) .

Taking $𝑐 = - 1$ gives $𝐷 (- 𝐴) =$ ${(- 1)}^{𝑛} 𝐷 (𝐴)$ . If the $𝑗$ -th column of $𝐴$ is $𝟎$ , then $𝟎 = 0 \cdot 𝟎$ , and linearity in column $𝑗$ gives $𝐷 (𝐴) = 0 \cdot 𝐷 (𝐴) = 0$ . (Equivalently, the linear map $𝐱 \mapsto 𝐷 (\dots, 𝐱, \dots)$ sends $𝟎$ to $0$ by Proposition 2.1.4.)

Remark 5.1.6. We state the axioms for columns because the columns of $𝐴$ are the vectors $𝐴 𝐞_{𝑗}$ , so they describe what $𝑇_{𝐴}$ does to the standard basis. The same axioms for rows would serve equally well, and Theorem 5.4.14 shows that they lead to the same function.

Multilinearity and alternation also make sense for a function $𝑓 : 𝑉 \times \dots \times 𝑉 \to 𝐹$ of $𝑘$ vectors in any vector space $𝑉$ . A multilinear such $𝑓$ is a $𝑘$ -linear form on $𝑉$ , and an alternating $𝑘$ -linear form if it is also alternating. The proofs of Proposition 5.1.10 and Proposition 5.1.17 below use nothing about columns of matrices, so those results hold for alternating $𝑘$ -linear forms with the same proofs. Section 13.4 develops this point of view and uses it to define the determinant of an operator without choosing a basis.

5.1.3Swapping columns

The axioms say nothing directly about what happens when two columns are interchanged. In the plane, interchanging $𝐚_{1}$ and $𝐚_{2}$ reverses the direction of the turn from one to the other, so we expect the sign of $𝐷$ to flip. Proving this is our first use of the basic move for multilinear functions.

Technique 5.1.7 (Expand by multilinearity). If an argument of a multilinear function is a sum, split it: linearity in slot $𝑗$ turns $𝐷 (\dots, \sum_{𝑘} 𝑐_{𝑘} 𝐱_{𝑘}, \dots)$ into $\sum_{𝑘} 𝑐_{𝑘} 𝐷 (\dots, 𝐱_{𝑘}, \dots)$ . Doing this in several slots turns one value of $𝐷$ into a sum over all ways of choosing one term in each slot, so $𝑚$ terms in each of $𝑟$ slots give $𝑚^{𝑟}$ terms. Usually the terms are standard basis vectors, or other vectors on which $𝐷$ is known, and the expansion reduces $𝐷 (𝐴)$ to finitely many known values. When $𝐷$ is also alternating, every term that uses the same vector in two slots vanishes, and typically most of the sum disappears.

Reach for it when you need a value of a multilinear function on general inputs and you know its values on special ones, much as a linear map is determined by its values on a basis (Technique 2.2.7). The canonical example is the $2 \times 2$ determinant (Theorem 5.1.13). In Section 5.3 the same expansion, carried out in all $𝑛$ columns, produces a formula for $𝐷$ on $𝐹^{𝑛 \times 𝑛}$ .

Used in 28 places and 5 practice exercises

Lemma 5.1.8, Proposition 5.1.10, Theorem 5.1.13, Proposition 5.1.17, Exercise 5.1.20, Exercise 5.1.33, Exercise 5.1.37, Exercise 5.1.38, Exercise 5.1.40, Exercise 5.2.44, Lemma 5.3.2, Theorem 5.3.3, Exercise 5.3.26, Theorem 5.5.3, Theorem 5.5.14, Theorem 5.5.27, Exercise 5.5.48, Exercise 5.5.52, Exercise 5.6.8, Problem 5.3, Problem 5.6, Proposition 6.1.22, Proposition 6.1.29, Exercise 6.1.44, Exercise 6.1.57, Corollary 6.3.6, Exercise 6.3.56, Exercise 7.3.53; 5 practice exercises.

The simplest expansion has just two terms, one of which vanishes.

Lemma 5.1.8 (Column replacement). Let

𝐷 : 𝐹^{𝑛 \times 𝑛} \to 𝐹

be multilinear and alternating, let

𝑖 \neq 𝑗

, and let

𝑐 \in 𝐹

. If

𝐴^{'}

is obtained from

𝐴

by adding

𝑐

times column

𝑗

to column

𝑖

, then

𝐷 (𝐴^{'}) = 𝐷 (𝐴)

Proof.routine

Uses. Definition 5.1.1 (Determinant function)

Techniques. Expand by multilinearity (Technique 5.1.7)

Hint 1. Column

𝑖

𝐴^{'}

is a sum,

𝐚_{𝑖} + 𝑐 𝐚_{𝑗}

. Expand

𝐷 (𝐴^{'})

in column

𝑖

Hint 2. One of the two terms is

𝐷 (𝐴)

. In the other, which vector appears in two columns?

Sketch. Linearity in column

𝑖

gives

𝐷 (𝐴^{'}) =

𝐷 (𝐴) +

𝑐 𝐷 (𝐴^{″})

, where

𝐴^{″}

𝐴

with column

𝑖

replaced by

𝐚_{𝑗}

. Then

𝐴^{″}

has equal columns

𝑖

and

𝑗

, so

𝐷 (𝐴^{″}) = 0

Let $𝐴^{″}$ be the matrix obtained from $𝐴$ by replacing column $𝑖$ with $𝐚_{𝑗}$ . The matrices $𝐴$ , $𝐴^{'}$ , and $𝐴^{″}$ agree outside column $𝑖$ , where they have $𝐚_{𝑖}$ , $𝐚_{𝑖} + 𝑐 𝐚_{𝑗}$ , and $𝐚_{𝑗}$ . By linearity in column $𝑖$ ,

𝐷 (𝐴^{'}) = 𝐷 (𝐴) + 𝑐 𝐷 (𝐴^{″}) .

Columns $𝑖$ and $𝑗$ of $𝐴^{″}$ are both $𝐚_{𝑗}$ , and $𝑖 \neq 𝑗$ , so $𝐷 (𝐴^{″}) = 0$ because $𝐷$ is alternating. Hence $𝐷 (𝐴^{'}) = 𝐷 (𝐴)$ . □

This is the algebraic form of the shear in Intuition (Signed area is linear in each column): sliding one edge of a parallelogram parallel to another edge does not change its signed area. Section 5.4 combines this lemma with the other column operations to compute determinants by elimination.

Definition 5.1.9 (Antisymmetric). A function

𝐷 : 𝐹^{𝑛 \times 𝑛} \to 𝐹

is antisymmetric if interchanging two columns negates its value: whenever

𝐴^{'}

is obtained from

𝐴

by interchanging columns

𝑖

and

𝑗

for some

𝑖 \neq 𝑗

, we have

𝐷 (𝐴^{'}) = - 𝐷 (𝐴)

For instance, $(\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix}) \mapsto$ $𝑎 𝑑 -$ $𝑏 𝑐$ is antisymmetric, since interchanging the columns gives $𝑏 𝑐 - 𝑎 𝑑$ . The function $(\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix}) \mapsto 𝑎 𝑑$ is not: interchanging the columns of $𝐼_{2}$ changes its value from $1$ to $0$ , not to $- 1$ .

Every result in this subsection fixes all columns except two, so we set up a shorthand. Fix $𝑖 \neq 𝑗$ and the columns $𝐚_{𝑘}$ for $𝑘 \notin {𝑖, 𝑗}$ , and for $𝐱, 𝐲 \in 𝐹^{𝑛}$ let $𝑓 (𝐱, 𝐲)$ be the value of $𝐷$ at the matrix with $𝐱$ in column $𝑖$ , $𝐲$ in column $𝑗$ , and $𝐚_{𝑘}$ in every other column $𝑘$ . If $𝐷$ is multilinear, then $𝑓$ is linear in each of $𝐱$ and $𝐲$ when the other is fixed; if $𝐷$ is alternating, then $𝑓 (𝐱, 𝐱) = 0$ ; and $𝐷$ is antisymmetric exactly when $𝑓 (𝐲, 𝐱) =$ $- 𝑓 (𝐱, 𝐲)$ for every such choice of $𝑖$ , $𝑗$ , and the fixed columns.

Proposition 5.1.10 (Alternating implies antisymmetric). If

𝐷 : 𝐹^{𝑛 \times 𝑛} \to 𝐹

is multilinear and alternating, then

𝐷

is antisymmetric.

Proof (by expanding).routine

Uses. Definition 5.1.1 (Determinant function), Definition 5.1.9 (Antisymmetric)

Techniques. Expand by multilinearity (Technique 5.1.7)

Hint 1. Only columns

𝑖

and

𝑗

change, so work with the two-variable function

𝑓

above. You want

𝑓 (𝐲, 𝐱) =

- 𝑓 (𝐱, 𝐲)

Hint 2. Alternation only says something when two columns are equal. Which matrix with two equal columns involves both

𝐱

and

𝐲

Hint 3. Expand

0 = 𝑓 (𝐱 + 𝐲, 𝐱 + 𝐲)

in both slots.

Sketch. Expanding

0 = 𝑓 (𝐱 + 𝐲, 𝐱 + 𝐲)

gives four terms. Alternation kills

𝑓 (𝐱, 𝐱)

and

𝑓 (𝐲, 𝐲)

, leaving

𝑓 (𝐱, 𝐲) +

𝑓 (𝐲, 𝐱) =

0

Fix $𝑖 \neq 𝑗$ and the other columns, and let $𝑓$ be as above. Let $𝐱, 𝐲 \in 𝐹^{𝑛}$ . Since $𝐷$ is alternating, $𝑓 (𝐱 + 𝐲, 𝐱 + 𝐲) = 0$ . Expanding by linearity in the first slot and then in the second,

\begin{aligned} 0 & = 𝑓 (𝐱, 𝐱 + 𝐲) + 𝑓 (𝐲, 𝐱 + 𝐲) \\ = 𝑓 (𝐱, 𝐱) + 𝑓 (𝐱, 𝐲) + 𝑓 (𝐲, 𝐱) + 𝑓 (𝐲, 𝐲) \\ = 𝑓 (𝐱, 𝐲) + 𝑓 (𝐲, 𝐱), \end{aligned}

where the last step uses $𝑓 (𝐱, 𝐱) =$ $𝑓 (𝐲, 𝐲) =$ $0$ , again because $𝐷$ is alternating. Hence $𝑓 (𝐲, 𝐱) =$ $- 𝑓 (𝐱, 𝐲)$ . As $𝑖$ , $𝑗$ , the other columns, $𝐱$ , and $𝐲$ were arbitrary, $𝐷$ is antisymmetric. □

The second proof avoids the four-term expansion. It performs the swap with moves whose effect on $𝐷$ we already know: three column replacements, which do not change $𝐷$ (Lemma 5.1.8), and one sign change. In the plane, the three replacements are shears, and together they rotate the pair $(𝐱, 𝐲)$ to $(𝐲, - 𝐱)$ .

Proof (by column replacements).routine

Uses. Lemma 5.1.8 (Column replacement), Definition 5.1.1 (Determinant function), Definition 5.1.9 (Antisymmetric)

Hint 1. Lemma 5.1.8 says that

𝑓 (𝐱 + 𝑐 𝐲, 𝐲) =

𝑓 (𝐱, 𝐲)

and

𝑓 (𝐱, 𝐲 + 𝑐 𝐱) =

𝑓 (𝐱, 𝐲)

. Try to reach

𝑓 (𝐲, \pm 𝐱)

by such moves.

Hint 2. Start with

𝑓 (𝐱, 𝐲) =

𝑓 (𝐱 + 𝐲, 𝐲)

. Now subtract the first slot from the second.

Sketch. Three column replacements take

(𝐱, 𝐲)

(𝐱 + 𝐲, 𝐲)

, then to

(𝐱 + 𝐲, - 𝐱)

, then to

(𝐲, - 𝐱)

. None of them changes

𝑓

, and pulling out the factor

- 1

gives

𝑓 (𝐱, 𝐲) =

- 𝑓 (𝐲, 𝐱)

Fix $𝑖 \neq 𝑗$ and the other columns, let $𝑓$ be as above, and let $𝐱, 𝐲 \in 𝐹^{𝑛}$ . Each of the first three steps below adds a multiple of one slot to the other, so it does not change $𝑓$ by Lemma 5.1.8:

\begin{aligned} 𝑓 (𝐱, 𝐲) & = 𝑓 (𝐱 + 𝐲, 𝐲) & (add slot 2 to slot 1) \\ = 𝑓 (𝐱 + 𝐲, 𝐲 - (𝐱 + 𝐲)) = 𝑓 (𝐱 + 𝐲, - 𝐱) & (subtract slot 1 from slot 2) \\ = 𝑓 (𝐱 + 𝐲 + (- 𝐱), - 𝐱) = 𝑓 (𝐲, - 𝐱) & (add slot 2 to slot 1) \\ = - 𝑓 (𝐲, 𝐱) & (linearity in slot 2) . \end{aligned}

So $𝑓 (𝐲, 𝐱) =$ $- 𝑓 (𝐱, 𝐲)$ , and $𝐷$ is antisymmetric. □

The first proof is the more economical one. The second shows that a swap is built from shears and a single reflection, which is how Section 5.4 analyzes column operations.

Antisymmetry looks like a reasonable alternative axiom: it is exactly the orientation reversal we observed in the plane. For multilinear functions the two conditions are equivalent over most fields, by Proposition 5.1.11 below. But the converse fails over fields in which $1 + 1 = 0$ (as soon as $𝑛 \geq 2$ ), and that failure is why we did not choose antisymmetry.

Proposition 5.1.11 (Antisymmetric implies alternating when

1 + 1 \neq 0

). Suppose

1 + 1 \neq 0

𝐹

. If

𝐷 : 𝐹^{𝑛 \times 𝑛} \to 𝐹

is antisymmetric, then

𝐷

is alternating. Consequently, a multilinear function on

𝐹^{𝑛 \times 𝑛}

is alternating if and only if it is antisymmetric.

Proof.routine

Uses. Definition 5.1.9 (Antisymmetric), Proposition 1.1.17 (No zero divisors), Proposition 5.1.10 (Alternating implies antisymmetric)

Hint 1. Suppose columns

𝑖

and

𝑗

𝐴

are equal. What does interchanging them do to

𝐴

Hint 2. It does not change

𝐴

, so antisymmetry gives

𝐷 (𝐴) = - 𝐷 (𝐴)

. Rewrite this as

(1 + 1) 𝐷 (𝐴) = 0

Sketch. If

𝐚_{𝑖} = 𝐚_{𝑗}

with

𝑖 \neq 𝑗

, interchanging these columns leaves

𝐴

unchanged, so

𝐷 (𝐴) = - 𝐷 (𝐴)

and

(1 + 1) 𝐷 (𝐴) = 0

. Since

1 + 1 \neq 0

𝐷 (𝐴) = 0

. The “consequently” combines this with Proposition 5.1.10.

Let $𝐴 \in 𝐹^{𝑛 \times 𝑛}$ with $𝐚_{𝑖} = 𝐚_{𝑗}$ for some $𝑖 \neq 𝑗$ . Interchanging columns $𝑖$ and $𝑗$ of $𝐴$ gives $𝐴$ again, so antisymmetry gives $𝐷 (𝐴) = - 𝐷 (𝐴)$ . Adding $𝐷 (𝐴)$ to both sides, $𝐷 (𝐴) + 𝐷 (𝐴) = 0$ , that is, $(1 + 1) 𝐷 (𝐴) = 0$ by distributivity. Since $1 + 1 \neq 0$ and a field has no zero divisors (Proposition 1.1.17), $𝐷 (𝐴) = 0$ . So $𝐷$ is alternating.

For the last sentence: a multilinear alternating function is antisymmetric by Proposition 5.1.10, and an antisymmetric function is alternating by what we just proved. □

Note that this direction does not use multilinearity at all. The condition $1 + 1 \neq 0$ says that $char 𝐹 \neq 2$ (Definition 1.1.28). It holds over $ℚ$ , $ℝ$ , $ℂ$ , and $𝔽_{𝑝}$ for odd $𝑝$ , and fails over $𝔽_{2}$ , where $- 1 = 1$ and antisymmetry says only that interchanging two columns changes nothing.

Example 5.1.12 (Antisymmetric but not alternating, over $𝔽_{2}$ ). On $𝔽_{2}^{2 \times 2}$ , define

𝐷 (\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix}) = 𝑎 𝑑 + 𝑏 𝑐 + 𝑎 𝑏 .

For a fixed second column $(𝑏, 𝑑)$ , the map $(𝑎, 𝑐) \mapsto$ $𝑎 𝑑 +$ $𝑏 𝑐 +$ $𝑎 𝑏 =$ $(𝑑 + 𝑏) 𝑎 +$ $𝑏 𝑐$ is linear, and for a fixed first column, $(𝑏, 𝑑) \mapsto$ $(𝑐 + 𝑎) 𝑏 +$ $𝑎 𝑑$ is linear, so $𝐷$ is multilinear. Interchanging the columns gives $𝑏 𝑐 + 𝑎 𝑑 + 𝑏 𝑎$ , the same value, and since $- 1 = 1$ in $𝔽_{2}$ this means $𝐷$ is antisymmetric. Also $𝐷 (𝐼_{2}) = 1 + 0 + 0 = 1$ . But $𝐷$ is not alternating, since $𝐷 (\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix}) = 0 + 0 + 1 = 1$ .

So over $𝔽_{2}$ the axioms “normalized, antisymmetric, multilinear” are satisfied by at least two different functions: this $𝐷$ , and $𝑎 𝑑 -$ $𝑏 𝑐 =$ $𝑎 𝑑 +$ $𝑏 𝑐$ , which we will see is a determinant function (Theorem 5.1.13). With alternation in place of antisymmetry, the $2 \times 2$ case has exactly one solution over every field. This is the reason the definition uses alternation, and an instance of the warning in Aside (Characteristic) that arguments dividing by $1 + 1$ fail in characteristic $2$ .

5.1.4The $2 \times 2$ determinant

We now show that the axioms determine $𝐷$ completely on $𝐹^{2 \times 2}$ .

Intuition (Strategy). The axioms tell us the value of

𝐷

on very few matrices:

𝐷 (𝐞_{1}, 𝐞_{1}) =

𝐷 (𝐞_{2}, 𝐞_{2}) =

0

by alternation,

𝐷 (𝐞_{2}, 𝐞_{1}) =

- 𝐷 (𝐞_{1}, 𝐞_{2})

by Proposition 5.1.10, and

𝐷 (𝐞_{1}, 𝐞_{2}) =

𝐷 (𝐼_{2}) =

1

by normalization. Multilinearity lets us reach every other matrix from these four. Write each column as a combination of

𝐞_{1}

and

𝐞_{2}

and expand (Technique 5.1.7):

𝐷 (𝐴)

becomes a combination of the four known values, with coefficients that are products of entries of

𝐴

. Since the expansion uses only multilinearity and alternation, we keep

𝐷 (𝐼_{2})

as an unknown until the end; this gives a stronger statement for free.

Theorem 5.1.13 (The $2 \times 2$ determinant).

(a)Let $𝐷 : 𝐹^{2 \times 2} \to 𝐹$ be multilinear and alternating. Then for all $𝑎, 𝑏, 𝑐, 𝑑 \in 𝐹$ ,
$𝐷 (\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix}) = (𝑎 𝑑 - 𝑏 𝑐) 𝐷 (𝐼_{2}) .$
(b)The function $(\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix}) \mapsto$ $𝑎 𝑑 -$ $𝑏 𝑐$ is a determinant function on $𝐹^{2 \times 2}$ .

Consequently, there is exactly one determinant function on $𝐹^{2 \times 2}$ , namely $(\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix}) \mapsto$ $𝑎 𝑑 -$ $𝑏 𝑐$ .

Proof, part 1 of 2: the axioms force $𝑎 𝑑 - 𝑏 𝑐$ .standard

Uses. Definition 5.1.1 (Determinant function), Proposition 5.1.10 (Alternating implies antisymmetric), Definition 1.6.8 (Standard bases)

Techniques. Expand by multilinearity (Technique 5.1.7)

Hint 1. Write each column of

𝐴

as a linear combination of

𝐞_{1}

and

𝐞_{2}

Hint 2. Expand

𝐷 (𝑎 𝐞_{1} + 𝑐 𝐞_{2}, 𝑏 𝐞_{1} + 𝑑 𝐞_{2})

by multilinearity, first in column

1

, then in column

2

. You get four terms.

Hint 3. The four terms involve

𝐷 (𝐞_{1}, 𝐞_{1})

𝐷 (𝐞_{1}, 𝐞_{2})

𝐷 (𝐞_{2}, 𝐞_{1})

, and

𝐷 (𝐞_{2}, 𝐞_{2})

. Evaluate them with alternation and Proposition 5.1.10.

Sketch. The columns are

𝑎 𝐞_{1} + 𝑐 𝐞_{2}

and

𝑏 𝐞_{1} + 𝑑 𝐞_{2}

. Expanding gives

𝑎 𝑏 𝐷 (𝐞_{1}, 𝐞_{1}) +

𝑎 𝑑 𝐷 (𝐞_{1}, 𝐞_{2}) +

𝑐 𝑏 𝐷 (𝐞_{2}, 𝐞_{1}) +

𝑐 𝑑 𝐷 (𝐞_{2}, 𝐞_{2})

. Alternation kills the first and last terms, and antisymmetry turns

𝐷 (𝐞_{2}, 𝐞_{1})

into

- 𝐷 (𝐞_{1}, 𝐞_{2}) =

- 𝐷 (𝐼_{2})

(a) Let $𝐴 = (\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix})$ . Its columns are $𝑎 𝐞_{1} + 𝑐 𝐞_{2}$ and $𝑏 𝐞_{1} + 𝑑 𝐞_{2}$ . By linearity in the first column,

𝐷 (𝐴) = 𝑎 𝐷 (𝐞_{1}, 𝑏 𝐞_{1} + 𝑑 𝐞_{2}) + 𝑐 𝐷 (𝐞_{2}, 𝑏 𝐞_{1} + 𝑑 𝐞_{2}),

and by linearity in the second column,

𝐷 (𝐴) = 𝑎 𝑏 𝐷 (𝐞_{1}, 𝐞_{1}) + 𝑎 𝑑 𝐷 (𝐞_{1}, 𝐞_{2}) + 𝑐 𝑏 𝐷 (𝐞_{2}, 𝐞_{1}) + 𝑐 𝑑 𝐷 (𝐞_{2}, 𝐞_{2}) .

Since $𝐷$ is alternating, $𝐷 (𝐞_{1}, 𝐞_{1}) =$ $𝐷 (𝐞_{2}, 𝐞_{2}) =$ $0$ . By Proposition 5.1.10, $𝐷 (𝐞_{2}, 𝐞_{1}) =$ $- 𝐷 (𝐞_{1}, 𝐞_{2})$ , and $𝐷 (𝐞_{1}, 𝐞_{2}) =$ $𝐷 (𝐼_{2})$ . Therefore

𝐷 (𝐴) = 𝑎 𝑑 𝐷 (𝐼_{2}) - 𝑏 𝑐 𝐷 (𝐼_{2}) = (𝑎 𝑑 - 𝑏 𝑐) 𝐷 (𝐼_{2}),

as desired.

Part (a) shows that there is at most one determinant function on $𝐹^{2 \times 2}$ : if $𝐷 (𝐼_{2}) = 1$ , then $𝐷 (𝐴) = 𝑎 𝑑 - 𝑏 𝑐$ . It remains to check that this candidate satisfies the axioms. For $𝑛 = 2$ this is a direct computation. For general $𝑛$ the analogous check is the existence half of Theorem 5.3.4, and it is where permutations enter.

Proof, part 2 of 2: $𝑎 𝑑 - 𝑏 𝑐$ satisfies the axioms.routine

Uses. Definition 5.1.1 (Determinant function), Definition 2.1.1 (Linear map)

Hint 1. Check the three axioms separately. For multilinearity, fix one column and look at

𝑎 𝑑 - 𝑏 𝑐

as a function of the entries of the other column.

Hint 2. For a fixed second column

(𝑏, 𝑑)

, the value is

𝑑 𝑎 - 𝑏 𝑐

, which is a fixed linear combination of the entries

𝑎

and

𝑐

of the first column.

Sketch. For a fixed second column,

(𝑎, 𝑐) \mapsto 𝑑 𝑎 - 𝑏 𝑐

is linear; for a fixed first column,

(𝑏, 𝑑) \mapsto 𝑎 𝑑 - 𝑐 𝑏

is linear. Equal columns give

𝑎 𝑐 - 𝑐 𝑎 = 0

, and

𝐼_{2}

gives

1 \cdot 1 - 0 \cdot 0 = 1

(b) Let $Δ (\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix}) =$ $𝑎 𝑑 -$ $𝑏 𝑐$ .

Normalized. $Δ (𝐼_{2}) = 1 \cdot 1 - 0 \cdot 0 = 1$ .

Alternating. If the two columns are equal, then $𝑏 = 𝑎$ and $𝑑 = 𝑐$ , so $Δ = 𝑎 𝑐 - 𝑎 𝑐 = 0$ .

Multilinear. Fix the second column $(𝑏, 𝑑)$ . For first columns $(𝑎, 𝑐)$ and $(𝑎^{'}, 𝑐^{'})$ and $𝑠 \in 𝐹$ ,

\begin{aligned} Δ (\begin{matrix} 𝑠 𝑎 + 𝑎^{'} & 𝑏 \\ 𝑠 𝑐 + 𝑐^{'} & 𝑑 \end{matrix}) & = (𝑠 𝑎 + 𝑎^{'}) 𝑑 - 𝑏 (𝑠 𝑐 + 𝑐^{'}) \\ = 𝑠 (𝑎 𝑑 - 𝑏 𝑐) + (𝑎^{'} 𝑑 - 𝑏 𝑐^{'}) = 𝑠 Δ (\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix}) + Δ (\begin{matrix} 𝑎^{'} & 𝑏 \\ 𝑐^{'} & 𝑑 \end{matrix}), \end{aligned}

so $Δ$ is linear in the first column. Similarly, for a fixed first column $(𝑎, 𝑐)$ ,

𝑎 (𝑠 𝑑 + 𝑑^{'}) - (𝑠 𝑏 + 𝑏^{'}) 𝑐 = 𝑠 (𝑎 𝑑 - 𝑏 𝑐) + (𝑎 𝑑^{'} - 𝑏^{'} 𝑐),

so $Δ$ is linear in the second column. Hence $Δ$ is a determinant function.

Finally, if $𝐷$ is any determinant function on $𝐹^{2 \times 2}$ , then $𝐷 (𝐼_{2}) = 1$ and (a) gives $𝐷 = Δ$ . So $Δ$ is the only one. □

Definition 5.1.14 (The

2 \times 2

determinant). For

𝐴 =

(\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix}) \in

𝐹^{2 \times 2}

, the determinant of

𝐴

det 𝐴 = 𝑎 𝑑 - 𝑏 𝑐

. By Theorem 5.1.13 it is the unique determinant function on

𝐹^{2 \times 2}

, and every multilinear alternating

𝐷 : 𝐹^{2 \times 2} \to 𝐹

is a scalar multiple of it:

𝐷 = 𝐷 (𝐼_{2}) det

For the matrices of Figure 5.1.1, $det 𝐴 =$ $2 \cdot$ $1 -$ $(- 1) \cdot$ $1 =$ $3$ , $det 𝑅 = - 1$ , and $det 𝐵 = 1 \cdot 2 - 2 \cdot 1 = 0$ , as the pictures suggested. In general, the formula is the signed area of the motivation.

Example 5.1.15 ( $det$ is signed area). Let $𝐚_{1}, 𝐚_{2} \in ℝ^{2}$ be nonzero, and write them in polar form, $𝐚_{1} =$ $𝑟 (cos 𝛼, sin 𝛼)$ and $𝐚_{2} =$ $𝑠 (cos 𝛽, sin 𝛽)$ with $𝑟, 𝑠 > 0$ . Then

det (𝐚_{1}, 𝐚_{2}) = 𝑟 𝑠 (cos 𝛼 sin 𝛽 - sin 𝛼 cos 𝛽) = 𝑟 𝑠 sin (𝛽 - 𝛼) .

The parallelogram $𝑃 (𝐚_{1}, 𝐚_{2})$ has base $𝑟$ and height $𝑠 | sin (𝛽 - 𝛼) |$ , so $| det (𝐚_{1}, 𝐚_{2}) |$ is its area. Also $sin (𝛽 - 𝛼) > 0$ exactly when $𝛽 - 𝛼$ lies strictly between $0$ and $𝜋$ modulo $2 𝜋$ , that is, when $𝐚_{2}$ is reached from $𝐚_{1}$ by a counterclockwise turn of less than a half-turn. So $det (𝐚_{1}, 𝐚_{2})$ is the signed area $𝑆 (𝐚_{1}, 𝐚_{2})$ of Intuition (Signed area is linear in each column). (If $𝐚_{1}$ or $𝐚_{2}$ is $𝟎$ , both are $0$ .) Together with part (b) of Theorem 5.1.13, this proves the three properties of signed area that we read off from pictures. The general statement that $| det 𝐴 |$ is the factor by which $𝑇_{𝐴}$ scales volume needs a theory of volume; see Section 9.2.

Exercise 5.1.16.routine Let

𝐴 = (\begin{matrix} 3 & 5 \\ 4 & 2 \end{matrix})

. Compute

det 𝐴

when

𝐴

is regarded as a matrix over

ℚ

, over

𝔽_{7}

, and over

𝔽_{2}

(reading each integer as its remainder). For which primes

𝑝

det 𝐴 = 0

over

𝔽_{𝑝}

Solution.

Uses. Definition 5.1.14 (The $2 \times 2$ determinant), Definition 1.1.19 (Integers modulo $𝑛$ ), Lemma 1.1.21 (Arithmetic in $ℤ / 𝑛 ℤ$ )

Hint 1. Over

ℚ

det 𝐴 = 3 \cdot 2 - 5 \cdot 4

Hint 2. Reducing the entries modulo

𝑝

and then computing gives the same result as computing over

ℤ

and then reducing, by Lemma 1.1.21.

Sketch. Over

ℚ

det 𝐴 = - 14

. Over

𝔽_{𝑝}

it is

- 14 mod 𝑝

, which is

0

exactly when

𝑝 ∣ 14

, that is,

𝑝 = 2

𝑝 = 7

Over

ℚ

det 𝐴 = 3 \cdot 2 - 5 \cdot 4 = - 14

. By Lemma 1.1.21, computing

𝑎 𝑑 - 𝑏 𝑐

𝔽_{𝑝}

from the remainders of the entries gives the remainder of

- 14

modulo

𝑝

. In

𝔽_{7}

this is

0

, since

7 ∣ 14

. In

𝔽_{2}

the matrix is

(\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix})

, with determinant

1 \cdot 0 - 1 \cdot 0 = 0

, as it should be since

2 ∣ 14

. In general

det 𝐴 = 0

over

𝔽_{𝑝}

exactly when

𝑝 ∣ 14

, that is, for

𝑝 = 2

and

𝑝 = 7

Intuition (What the proof did, and where it leads). The expansion in Theorem 5.1.13 produced one term for each way of choosing $𝐞_{1}$ or $𝐞_{2}$ in each column. The two terms that chose the same vector twice died, and the two survivors correspond to the two orders of $𝐞_{1}, 𝐞_{2}$ : the identity order, with value $𝐷 (𝐼_{2})$ , and the swapped order, with value $- 𝐷 (𝐼_{2})$ .

For an $𝑛 \times 𝑛$ matrix, write column $𝑗$ as $\sum_{𝑖} 𝑎_{𝑖 𝑗} 𝐞_{𝑖}$ and expand all $𝑛$ columns. This gives $𝑛^{𝑛}$ terms, each a product of entries times some $𝐷 (𝐞_{𝑖_{1}}, \dots, 𝐞_{𝑖_{𝑛}})$ . Alternation kills every term with a repeated index, leaving one term for each ordering $𝑖_{1}, \dots, 𝑖_{𝑛}$ of $1, \dots, 𝑛$ . Each surviving value can be computed by interchanging columns until the order is $1, \dots, 𝑛$ , picking up a factor $- 1$ per interchange. For example, two interchanges sort $𝐞_{2}, 𝐞_{3}, 𝐞_{1}$ :

𝐷 (𝐞_{2}, 𝐞_{3}, 𝐞_{1}) = - 𝐷 (𝐞_{1}, 𝐞_{3}, 𝐞_{2}) = 𝐷 (𝐞_{1}, 𝐞_{2}, 𝐞_{3}) .

Two questions remain. First, could one ordering be sorted both by an even and by an odd number of interchanges? If so, the axioms would force $𝐷 (𝐼_{𝑛}) =$ $- 𝐷 (𝐼_{𝑛})$ , so $𝐷 (𝐼_{𝑛}) = 0$ when $1 + 1 \neq 0$ , and no determinant function could exist. Second, does the formula produced by the expansion actually satisfy the axioms? Section 5.2 answers the first question (no), and Section 5.3 answers the second (yes).

5.1.5Dependent columns

In Figure 5.1.1, the dependent columns of $𝐵$ span a flat parallelogram, and its signed area is $0$ . The same holds in every dimension: if the columns of $𝐴$ are linearly dependent, then one column is a combination of the others, and expanding that column (Technique 5.1.7) produces only terms with a repeated column.

Proposition 5.1.17 (Dependent columns). Let

𝐷 : 𝐹^{𝑛 \times 𝑛} \to 𝐹

be multilinear and alternating. If the columns of

𝐴 \in 𝐹^{𝑛 \times 𝑛}

are linearly dependent, then

𝐷 (𝐴) = 0

Proof (by expanding).routine

Uses. Definition 1.5.1 (Linear independence), Definition 5.1.1 (Determinant function), Exercise 5.1.5

Techniques. Expand by multilinearity (Technique 5.1.7)

Hint 1. Use a nontrivial dependence relation to write one column

𝐚_{𝑘}

as a linear combination of the other columns.

Hint 2. Expand

𝐷 (𝐴)

in column

𝑘

. Each term has some

𝐚_{𝑗}

with

𝑗 \neq 𝑘

in column

𝑘

. Where else does

𝐚_{𝑗}

appear?

Sketch. Some

𝐚_{𝑘}

equals

\sum_{𝑗 \neq 𝑘} 𝑏_{𝑗} 𝐚_{𝑗}

. Linearity in column

𝑘

gives

𝐷 (𝐴) =

\sum_{𝑗 \neq 𝑘} 𝑏_{𝑗} 𝐷 (𝐴_{𝑗})

, where

𝐴_{𝑗}

has

𝐚_{𝑗}

in columns

𝑗

and

𝑘

. Each

𝐷 (𝐴_{𝑗})

0

Since the columns are linearly dependent, there are scalars $𝑐_{1}, \dots, 𝑐_{𝑛}$ , not all $0$ , with $𝑐_{1} 𝐚_{1} +$ $\dots +$ $𝑐_{𝑛} 𝐚_{𝑛} =$ $𝟎$ . Choose $𝑘$ with $𝑐_{𝑘} \neq 0$ . Then

𝐚_{𝑘} = \sum_{𝑗 \neq 𝑘} 𝑏_{𝑗} 𝐚_{𝑗}, where 𝑏_{𝑗} = - 𝑐_{𝑘}^{- 1} 𝑐_{𝑗} .

If $𝑛 = 1$ , the sum is empty, so $𝐚_{1} = 𝟎$ and $𝐷 (𝐴) = 0$ by Exercise 5.1.5. Otherwise, for $𝑗 \neq 𝑘$ let $𝐴_{𝑗}$ be the matrix obtained from $𝐴$ by replacing column $𝑘$ with $𝐚_{𝑗}$ . By linearity in column $𝑘$ ,

𝐷 (𝐴) = \sum_{𝑗 \neq 𝑘} 𝑏_{𝑗} 𝐷 (𝐴_{𝑗}) .

For each $𝑗 \neq 𝑘$ , columns $𝑗$ and $𝑘$ of $𝐴_{𝑗}$ are both $𝐚_{𝑗}$ , so $𝐷 (𝐴_{𝑗}) = 0$ because $𝐷$ is alternating. Hence $𝐷 (𝐴) = 0$ . □

The second proof replaces the expansion by an invariance argument: change $𝐴$ , without changing $𝐷 (𝐴)$ , until $𝐷$ is obviously $0$ .

Proof (by column replacement).routine

Uses. Definition 1.5.1 (Linear independence), Lemma 5.1.8 (Column replacement), Exercise 5.1.5

Hint 1. Again write

𝐚_{𝑘} =

\sum_{𝑗 \neq 𝑘} 𝑏_{𝑗} 𝐚_{𝑗}

Hint 2. Use Lemma 5.1.8 to subtract

𝑏_{𝑗} 𝐚_{𝑗}

from column

𝑘

, one

𝑗

at a time. What is column

𝑘

at the end?

Sketch. Subtracting

𝑏_{𝑗} 𝐚_{𝑗}

from column

𝑘

for each

𝑗 \neq 𝑘

does not change

𝐷

and turns column

𝑘

into

𝟎

. A matrix with a zero column has

𝐷 = 0

As in the first proof, write

𝐚_{𝑘} =

\sum_{𝑗 \neq 𝑘} 𝑏_{𝑗} 𝐚_{𝑗}

. For each

𝑗 \neq 𝑘

in turn, add

- 𝑏_{𝑗}

times column

𝑗

to column

𝑘

. Columns other than

𝑘

never change, so each step adds

- 𝑏_{𝑗} 𝐚_{𝑗}

to column

𝑘

, and by Lemma 5.1.8 it does not change the value of

𝐷

. After all the steps, column

𝑘

𝐚_{𝑘} -

\sum_{𝑗 \neq 𝑘} 𝑏_{𝑗} 𝐚_{𝑗} =

𝟎

. So

𝐷 (𝐴)

equals the value of

𝐷

at a matrix with a zero column, which is

0

by Exercise 5.1.5. □

The first proof is self-contained. The second generalizes better in practice: computing $𝐷 (𝐴)$ by column operations that either preserve $𝐷$ or change it in a known way, until the matrix is simple, is how determinants are evaluated in Section 5.4.

Corollary 5.1.18 (Non-invertible matrices). Let

𝐷 : 𝐹^{𝑛 \times 𝑛} \to 𝐹

be multilinear and alternating. If

𝐴 \in 𝐹^{𝑛 \times 𝑛}

is not invertible, then

𝐷 (𝐴) = 0

. Equivalently, if

𝐷 (𝐴) \neq 0

, then

𝐴

is invertible.

Proof.routine

Uses. Proposition 5.1.17 (Dependent columns), Corollary 3.5.25 (The invertible matrix theorem), Theorem 3.5.12 (First Fundamental Theorem, completed)

Hint 1. Relate invertibility of a square matrix to linear independence of its columns.

Hint 2. For

𝐴 \in 𝐹^{𝑛 \times 𝑛}

, Theorem 3.5.12 says the columns are independent if and only if

rank 𝐴 = 𝑛

, and Corollary 3.5.25 says

𝐴

is invertible if and only if

rank 𝐴 = 𝑛

Sketch. If

𝐴

is not invertible, then

rank 𝐴 \neq 𝑛

(Corollary 3.5.25), so the columns of

𝐴

are dependent (Theorem 3.5.12), and

𝐷 (𝐴) = 0

by Proposition 5.1.17.

Suppose

𝐴 \in 𝐹^{𝑛 \times 𝑛}

is not invertible. By Corollary 3.5.25,

rank 𝐴 \neq 𝑛

. By Theorem 3.5.12, the columns of

𝐴

are then not linearly independent. So

𝐷 (𝐴) = 0

by Proposition 5.1.17. The second statement is the contrapositive of the first. □

The converse is false for $𝐷 = 0$ , which is multilinear and alternating but vanishes on every matrix. So any proof that $det 𝐴 \neq 0$ for invertible $𝐴$ must use normalization. For $𝑛 = 2$ it follows from the formula (Exercise 5.1.25), and for general $𝑛$ it is Theorem 5.4.7.

Exercises

Exercise 5.1.19.routine For each function $𝐷 : ℚ^{2 \times 2} \to ℚ$ below, where $𝐴 = (\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix})$ , decide which of the three determinant axioms it satisfies. Then decide which of your answers change if $ℚ$ is replaced by $𝔽_{2}$ .

(a) $𝐷 (𝐴) = 𝑎 + 𝑑$ .
(b) $𝐷 (𝐴) = 𝑎 𝑏 - 𝑐 𝑑$ .
(c) $𝐷 (𝐴) = 𝑏 𝑐 - 𝑎 𝑑$ .
(d) $𝐷 (𝐴) =$ $(𝑎 - 𝑏) (𝑑 - 𝑐)$ .

Solution.

Uses. Definition 5.1.1 (Determinant function), Theorem 5.1.13 (The $2 \times 2$ determinant), Proposition 2.1.4 (Linear maps preserve $𝟎$ and negatives)

Hint 1. Normalization and alternation are a single evaluation and a substitution

𝑏 = 𝑎

𝑑 = 𝑐

. For multilinearity, fix one column and ask whether the value is a linear function of the entries of the other column.

Hint 2. A nonzero constant term, or a product of two entries from the same column, prevents linearity in that column.

Hint 3. Over

𝔽_{2}

- 1 = 1

. Which of the four functions is affected?

Sketch. Over

ℚ

: (a) fails all three axioms; (b) is multilinear only; (c) is

- det

, which is multilinear and alternating but not normalized; (d) is normalized and alternating, but contains the product

𝑎 𝑐

of two entries of the first column, so it is not multilinear. Over

𝔽_{2}

only (c) changes: there

- det = det

is a determinant function.

(a) $𝐷 (𝐼_{2}) = 2 \neq 1$ and $𝐷 (\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix}) = 1 \neq 0$ , so $𝐷$ is neither normalized nor alternating. For the fixed second column $(0, 1)$ , the map $(𝑎, 𝑐) \mapsto 𝑎 + 1$ sends $𝟎$ to $1$ , so it is not linear (Proposition 2.1.4), and $𝐷$ is not multilinear.

(b) For a fixed second column $(𝑏, 𝑑)$ , $(𝑎, 𝑐) \mapsto 𝑏 𝑎 - 𝑑 𝑐$ is linear, and for a fixed first column $(𝑎, 𝑐)$ , $(𝑏, 𝑑) \mapsto 𝑎 𝑏 - 𝑐 𝑑$ is linear, so $𝐷$ is multilinear. But $𝐷 (𝐼_{2}) = 1 \cdot 0 - 0 \cdot 1 = 0$ and $𝐷 (\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix}) = 1$ , so $𝐷$ is neither normalized nor alternating.

(c) $𝐷 = - det$ . By Theorem 5.1.13, $det$ is multilinear and alternating, and negating both sides of each defining identity shows that $- det$ is too. But $𝐷 (𝐼_{2}) = - 1 \neq 1$ .

(d) $𝐷 (𝐼_{2}) =$ $(1 - 0) (1 - 0) =$ $1$ , and if the columns are equal, then $𝑎 = 𝑏$ and $𝐷 (𝐴) = 0$ . So $𝐷$ is normalized and alternating. For the fixed second column $(0, 0)$ , $𝐷 (𝐴) =$ $𝑎 (- 𝑐) =$ $- 𝑎 𝑐$ . The map $(𝑎, 𝑐) \mapsto - 𝑎 𝑐$ is not linear: it sends $(1, 1)$ to $- 1$ and $(2, 2) = 2 (1, 1)$ to $- 4 \neq 2 \cdot (- 1)$ . So $𝐷$ is not multilinear.

Over $𝔽_{2}$ . In (a), $𝐷 (𝐼_{2}) = 1 + 1 = 0$ , and the other two failures are unchanged. In (b) nothing changes. In (d), the map $(𝑎, 𝑐) \mapsto$ $- 𝑎 𝑐 =$ $𝑎 𝑐$ sends $(1, 0)$ and $(0, 1)$ to $0$ but their sum $(1, 1)$ to $1$ , so it is still not linear. In (c), $- 1 = 1$ , so $𝐷 = - det = det$ is a determinant function. So only the answer to (c) changes.

Exercise 5.1.20.routine Suppose $𝐷$ is a determinant function on $𝐹^{3 \times 3}$ . Using only the axioms and the results of this section, find $𝐷 (𝐴)$ for

(a) 𝐴 = (\begin{matrix} 0 & 0 & 2 \\ 0 & 3 & 0 \\ 5 & 0 & 0 \end{matrix}), (b) 𝐴 = (\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix}), (c) 𝐴 = (\begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{matrix}) .

Solution.

Uses. Definition 5.1.1 (Determinant function), Proposition 5.1.10 (Alternating implies antisymmetric), Lemma 5.1.8 (Column replacement)

Techniques. Expand by multilinearity (Technique 5.1.7)

Hint 1. In (a) and (b), every column is a multiple of a standard basis vector. Pull out the scalars, then sort the columns into the order

𝐞_{1}, 𝐞_{2}, 𝐞_{3}

by interchanges.

Hint 2. In (c), use Lemma 5.1.8 to clear the entries above the diagonal, working from the left.

Sketch. (a)

𝐷 =

5 \cdot

3 \cdot

2 \cdot

𝐷 (𝐞_{3}, 𝐞_{2}, 𝐞_{1}) =

- 30

, one interchange. (b)

𝐷 (𝐞_{3}, 𝐞_{1}, 𝐞_{2}) = 1

, two interchanges. (c) Column replacements turn

𝐴

into

𝐼_{3}

, so

𝐷 (𝐴) = 1

(a) The columns are $5 𝐞_{3}$ , $3 𝐞_{2}$ , and $2 𝐞_{1}$ . By linearity in each column and then Proposition 5.1.10 (interchanging columns $1$ and $3$ ),

𝐷 (𝐴) = 30 𝐷 (𝐞_{3}, 𝐞_{2}, 𝐞_{1}) = - 30 𝐷 (𝐞_{1}, 𝐞_{2}, 𝐞_{3}) = - 30 .

(b) The columns are $𝐞_{3}, 𝐞_{1}, 𝐞_{2}$ . Interchanging columns $1$ and $2$ , then columns $2$ and $3$ ,

𝐷 (𝐴) = 𝐷 (𝐞_{3}, 𝐞_{1}, 𝐞_{2}) = - 𝐷 (𝐞_{1}, 𝐞_{3}, 𝐞_{2}) = 𝐷 (𝐞_{1}, 𝐞_{2}, 𝐞_{3}) = 1 .

(c) The columns are $𝐞_{1}$ , $2 𝐞_{1} + 𝐞_{2}$ , and $3 𝐞_{1} + 4 𝐞_{2} + 𝐞_{3}$ . By Lemma 5.1.8, subtracting $2$ times column $1$ from column $2$ , then $3$ times column $1$ from column $3$ , does not change $𝐷$ , and leaves the columns $𝐞_{1}$ , $𝐞_{2}$ , $4 𝐞_{2} + 𝐞_{3}$ . Subtracting $4$ times the new column $2$ from column $3$ leaves $𝐞_{1}, 𝐞_{2}, 𝐞_{3}$ . So $𝐷 (𝐴) = 𝐷 (𝐼_{3}) = 1$ .

Exercise 5.1.21.routine Find the error in the following argument.

Claim. If $𝐷$ is a determinant function on $𝐹^{𝑛 \times 𝑛}$ , then $𝐷 (𝐴 + 𝐵) =$ $𝐷 (𝐴) +$ $𝐷 (𝐵)$ for all $𝐴, 𝐵 \in 𝐹^{𝑛 \times 𝑛}$ .

Proof. Column $𝑗$ of $𝐴 + 𝐵$ is $𝐚_{𝑗} + 𝐛_{𝑗}$ . Since $𝐷$ is linear in each column,

𝐷 (𝐴 + 𝐵) = 𝐷 (𝐚_{1} + 𝐛_{1}, \dots, 𝐚_{𝑛} + 𝐛_{𝑛}) = 𝐷 (𝐚_{1}, \dots, 𝐚_{𝑛}) + 𝐷 (𝐛_{1}, \dots, 𝐛_{𝑛}) = 𝐷 (𝐴) + 𝐷 (𝐵) .

Solution.

Uses. Definition 5.1.1 (Determinant function), Exercise 5.1.5, Remark 5.1.4

Hint 1. Linearity in column

𝑗

splits column

𝑗

while the other columns stay fixed. Carry out the splitting honestly for

𝑛 = 2

Hint 2. Test the claim with

𝐴 = (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix})

and

𝐵 = (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix})

Sketch. For

𝑛 \geq 2

, splitting each column separately produces

2^{𝑛}

terms, not

2

; the proof keeps only two of them. For

𝐴 = (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix})

and

𝐵 = (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix})

det (𝐴 + 𝐵) = 1

but

det 𝐴 + det 𝐵 = 0

The middle equality is wrong. Linearity in column $𝑗$ applies when the other columns are held fixed, so it splits one column at a time. For $𝑛 = 2$ , splitting column $1$ and then column $2$ gives four terms:

𝐷 (𝐴 + 𝐵) = 𝐷 (𝐚_{1}, 𝐚_{2}) + 𝐷 (𝐚_{1}, 𝐛_{2}) + 𝐷 (𝐛_{1}, 𝐚_{2}) + 𝐷 (𝐛_{1}, 𝐛_{2}),

and in general there are $2^{𝑛}$ terms. The argument silently drops all of them except $𝐷 (𝐴)$ and $𝐷 (𝐵)$ . The claim is false for every $𝑛 \geq 2$ . For $𝑛 = 2$ , take $𝐴 = (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix})$ and $𝐵 = (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix})$ : then $𝐴 + 𝐵 = 𝐼_{2}$ , so $𝐷 (𝐴 + 𝐵) = 1$ , while $𝐷 (𝐴) = 𝐷 (𝐵) = 0$ because each has a zero column (Exercise 5.1.5). (For $𝑛 = 1$ the claim is true, since $𝐷 (𝑎) = 𝑎$ by Example 5.1.2.)

Exercise 5.1.22.routine Find the error in the following argument.

Claim. If $𝐷 : 𝐹^{𝑛 \times 𝑛} \to 𝐹$ is multilinear and alternating and $𝐷 (𝐴) = 0$ , then the columns of $𝐴$ are linearly dependent.

Proof. By Proposition 5.1.17, a matrix with dependent columns has $𝐷 (𝐴) = 0$ . Here $𝐷 (𝐴) = 0$ , so the columns of $𝐴$ are dependent.

Solution.

Uses. Proposition 5.1.17 (Dependent columns), Theorem 5.1.13 (The $2 \times 2$ determinant)

Hint 1. Compare the logical form of the proposition with the way it is used.

Hint 2. Is there a multilinear alternating

𝐷

that vanishes at a matrix with independent columns?

Sketch. The argument uses the converse of Proposition 5.1.17. The claim is false for

𝐷 = 0

and

𝐴 = 𝐼_{𝑛}

Proposition 5.1.17 says “dependent columns $\Rightarrow$ $𝐷 (𝐴) = 0$ ”. The argument uses the converse, “ $𝐷 (𝐴) = 0$ $\Rightarrow$ dependent columns”, which does not follow. In fact the claim is false: $𝐷 = 0$ is multilinear and alternating, and $𝐷 (𝐼_{𝑛}) = 0$ although the columns of $𝐼_{𝑛}$ are independent.

For a determinant function, the conclusion is true, but the proof must use normalization: for $𝑛 = 2$ see Exercise 5.1.25, and for general $𝑛$ see Theorem 5.4.7.

Exercise 5.1.23.routine Let $𝑎, 𝑏, 𝑐, 𝑑 > 0$ be real numbers with $𝑎 𝑑 > 𝑏 𝑐$ . Figure 5.1.2 shows the parallelogram $𝑃$ spanned by $(𝑎, 𝑐)$ and $(𝑏, 𝑑)$ inside the rectangle $[0, 𝑎 + 𝑏] \times$ $[0, 𝑐 + 𝑑]$ .

Figure 5.1.2. The parallelogram spanned by

(𝑎, 𝑐)

and

(𝑏, 𝑑)

, drawn for

(𝑎, 𝑏, 𝑐, 𝑑) =

(3, 1, 1, 2)

. The rest of the rectangle splits into four right triangles and two rectangles.

(a)By computing the areas of the other pieces, show that $𝑃$ has area $𝑎 𝑑 - 𝑏 𝑐$ .
(b)Explain why the hypothesis $𝑎 𝑑 > 𝑏 𝑐$ means that the turn from $(𝑎, 𝑐)$ to $(𝑏, 𝑑)$ is counterclockwise, so that the answer agrees with the signed area $det (\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix})$ .

Solution.

Uses. Definition 5.1.14 (The $2 \times 2$ determinant), Example 5.1.15 ( $det$ is signed area)

Hint 1. The pieces are: two right triangles with legs

𝑎

and

𝑐

, two with legs

𝑏

and

𝑑

, and two rectangles with sides

𝑏

and

𝑐

Hint 2. For (b), compare the slopes

𝑐 / 𝑎

and

𝑑 / 𝑏

of the two edges.

Sketch. The area of

𝑃

(𝑎 + 𝑏) (𝑐 + 𝑑) -

𝑎 𝑐 -

𝑏 𝑑 -

2 𝑏 𝑐 =

𝑎 𝑑 -

𝑏 𝑐

. The inequality

𝑎 𝑑 > 𝑏 𝑐

says

𝑐 / 𝑎 < 𝑑 / 𝑏

, so

(𝑏, 𝑑)

is steeper than

(𝑎, 𝑐)

and lies counterclockwise from it.

(a) Because $𝑎 𝑑 > 𝑏 𝑐$ , the vector $(𝑎, 𝑐)$ has smaller slope than $(𝑏, 𝑑)$ (see (b)), so $𝑃$ sits in the rectangle as in Figure 5.1.2, and $𝑃$ together with the six pieces below tiles the rectangle without overlaps. (If $𝑎 𝑑 \leq 𝑏 𝑐$ , the pieces would overlap and the computation below would not apply.) The rectangle has area $(𝑎 + 𝑏) (𝑐 + 𝑑)$ . Outside $𝑃$ it contains two right triangles with legs $𝑎$ and $𝑐$ (below the edge from $𝟎$ to $(𝑎, 𝑐)$ and above the opposite edge), two right triangles with legs $𝑏$ and $𝑑$ (left of the edge from $𝟎$ to $(𝑏, 𝑑)$ and right of the opposite edge), and two rectangles with sides $𝑏$ and $𝑐$ (in the lower right and upper left corners). So

area (𝑃) = (𝑎 + 𝑏) (𝑐 + 𝑑) - 2 \cdot \frac{𝑎 𝑐}{2} - 2 \cdot \frac{𝑏 𝑑}{2} - 2 𝑏 𝑐 = 𝑎 𝑐 + 𝑎 𝑑 + 𝑏 𝑐 + 𝑏 𝑑 - 𝑎 𝑐 - 𝑏 𝑑 - 2 𝑏 𝑐 = 𝑎 𝑑 - 𝑏 𝑐 .

(b) Since $𝑎, 𝑏 > 0$ , dividing $𝑎 𝑑 > 𝑏 𝑐$ by $𝑎 𝑏$ gives $𝑑 / 𝑏 > 𝑐 / 𝑎$ . Both vectors lie in the open first quadrant, and $(𝑏, 𝑑)$ has the larger slope, so its polar angle is larger: turning from $(𝑎, 𝑐)$ to $(𝑏, 𝑑)$ is a counterclockwise turn through an angle in $(0, 𝜋 / 2)$ . So the signed area of Example 5.1.15 is $+ area (𝑃) =$ $𝑎 𝑑 -$ $𝑏 𝑐 =$ $det (\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix})$ , consistent with (a).

Exercise 5.1.24.routine The permanent of $𝐴 = (\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix})$ is $per 𝐴 = 𝑎 𝑑 + 𝑏 𝑐$ .

(a)Show that $per$ is normalized and multilinear, and that interchanging the two columns of $𝐴$ does not change $per 𝐴$ .
(b)For which fields $𝐹$ is $per$ a determinant function on $𝐹^{2 \times 2}$ ?

Solution.

Uses. Definition 5.1.1 (Determinant function), Definition 1.1.28 (Characteristic)

Hint 1. For (b), only alternation is in doubt. Compute

per

at a matrix with two equal columns

(𝑎, 𝑐)

Hint 2. You get

𝑎 𝑐 +

𝑐 𝑎 =

(1 + 1) 𝑎 𝑐

. When does this vanish for all

𝑎, 𝑐

Sketch. (a) is a direct check. (b) With equal columns,

per 𝐴 = (1 + 1) 𝑎 𝑐

, which vanishes for all

𝑎, 𝑐

exactly when

1 + 1 = 0

. So

per

is a determinant function exactly when

char 𝐹 = 2

, and then

per = det

(a) $per (𝐼_{2}) =$ $1 \cdot$ $1 +$ $0 \cdot$ $0 =$ $1$ . For a fixed second column $(𝑏, 𝑑)$ , the map $(𝑎, 𝑐) \mapsto 𝑑 𝑎 + 𝑏 𝑐$ is linear, and for a fixed first column $(𝑎, 𝑐)$ , the map $(𝑏, 𝑑) \mapsto 𝑐 𝑏 + 𝑎 𝑑$ is linear, so $per$ is multilinear. Interchanging the columns gives $(\begin{matrix} 𝑏 & 𝑎 \\ 𝑑 & 𝑐 \end{matrix})$ , whose permanent is $𝑏 𝑐 + 𝑎 𝑑 = per 𝐴$ .

(b) By (a), $per$ is a determinant function if and only if it is alternating. A matrix with equal columns is $(\begin{matrix} 𝑎 & 𝑎 \\ 𝑐 & 𝑐 \end{matrix})$ , with permanent $𝑎 𝑐 +$ $𝑎 𝑐 =$ $(1 + 1) 𝑎 𝑐$ . If $1 + 1 = 0$ , this is $0$ for all $𝑎, 𝑐$ . If $1 + 1 \neq 0$ , then $𝑎 = 𝑐 = 1$ gives $1 + 1 \neq 0$ . So $per$ is a determinant function on $𝐹^{2 \times 2}$ if and only if $1 + 1 = 0$ , that is, $char 𝐹 = 2$ (Definition 1.1.28). In that case $- 𝑏 𝑐 = 𝑏 𝑐$ , so $per = det$ , as Theorem 5.1.13 requires.

Exercise 5.1.25.standard Let $𝐴 =$ $(\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix}) \in$ $𝐹^{2 \times 2}$ .

(a)Show that $det 𝐴 = 0$ if and only if the columns of $𝐴$ are linearly dependent.
(b)Deduce that $𝐴$ is invertible if and only if $det 𝐴 \neq 0$ , and that in this case
$𝐴^{- 1} = \frac{1}{det 𝐴} (\begin{matrix} 𝑑 & - 𝑏 \\ - 𝑐 & 𝑎 \end{matrix}) .$

Solution.

Uses. Proposition 5.1.17 (Dependent columns), Definition 1.5.1 (Linear independence), Corollary 3.5.25 (The invertible matrix theorem), Theorem 3.5.12 (First Fundamental Theorem, completed), Definition 2.4.1 (Matrix product), Definition 2.5.12 (Invertible)

Hint 1. One direction is Proposition 5.1.17. For the other, you need an explicit dependence relation built from

𝑎, 𝑏, 𝑐, 𝑑

Hint 2. Compute

𝑑 (𝑎, 𝑐) -

𝑐 (𝑏, 𝑑)

and

- 𝑏 (𝑎, 𝑐) +

𝑎 (𝑏, 𝑑)

Hint 3. Both are multiples of

det 𝐴

. When are their coefficients not both zero?

Sketch.

𝑑 𝐚_{1} -

𝑐 𝐚_{2} =

(det 𝐴, 0)

and

- 𝑏 𝐚_{1} +

𝑎 𝐚_{2} =

(0, det 𝐴)

. If

det 𝐴 = 0

, one of these is a nontrivial dependence unless

𝐴 = 0

. For (b), combine with the Fundamental Theorems, and multiply out to check the inverse.

Write $𝐚_{1} = (𝑎, 𝑐)$ and $𝐚_{2} = (𝑏, 𝑑)$ .

(a) If the columns are dependent, then $det 𝐴 = 0$ by Proposition 5.1.17, since $det$ is multilinear and alternating (Theorem 5.1.13). Conversely, suppose $det 𝐴 = 0$ . We compute

𝑑 𝐚_{1} - 𝑐 𝐚_{2} = (𝑑 𝑎 - 𝑐 𝑏, 𝑑 𝑐 - 𝑐 𝑑) = (det 𝐴, 0) = 𝟎, - 𝑏 𝐚_{1} + 𝑎 𝐚_{2} = (- 𝑏 𝑎 + 𝑎 𝑏, - 𝑏 𝑐 + 𝑎 𝑑) = (0, det 𝐴) = 𝟎 .

If $(𝑐, 𝑑) \neq (0, 0)$ , the first relation has coefficients $𝑑, - 𝑐$ not both zero, so the columns are dependent. If $𝑐 = 𝑑 = 0$ but $(𝑎, 𝑏) \neq (0, 0)$ , the second relation has coefficients $- 𝑏, 𝑎$ not both zero. If $𝑎 = 𝑏 = 𝑐 = 𝑑 = 0$ , then $𝐚_{1} = 𝟎$ , and $1 \cdot 𝐚_{1} + 0 \cdot 𝐚_{2} = 𝟎$ is a dependence. In every case the columns are dependent.

(b) By Corollary 3.5.25 and Theorem 3.5.12 (as in the proof of Corollary 5.1.18), a square matrix is invertible if and only if its columns are linearly independent. By (a), this happens if and only if $det 𝐴 \neq 0$ . Now let $𝐶 = (\begin{matrix} 𝑑 & - 𝑏 \\ - 𝑐 & 𝑎 \end{matrix})$ . Multiplying out,

𝐴 𝐶 = (\begin{matrix} 𝑎 𝑑 - 𝑏 𝑐 & - 𝑎 𝑏 + 𝑏 𝑎 \\ 𝑐 𝑑 - 𝑑 𝑐 & - 𝑐 𝑏 + 𝑑 𝑎 \end{matrix}) = (det 𝐴) 𝐼_{2}, 𝐶 𝐴 = (\begin{matrix} 𝑑 𝑎 - 𝑏 𝑐 & 𝑑 𝑏 - 𝑏 𝑑 \\ - 𝑐 𝑎 + 𝑎 𝑐 & - 𝑐 𝑏 + 𝑎 𝑑 \end{matrix}) = (det 𝐴) 𝐼_{2} .

If $det 𝐴 \neq 0$ , it follows that ${(det 𝐴)}^{- 1} 𝐶$ is a two-sided inverse of $𝐴$ .

Exercise 5.1.26.routine

(a)Compute $det (\begin{matrix} 1 + 𝑖 & 2 \\ 𝑖 & 1 - 𝑖 \end{matrix})$ over $ℂ$ .
(b)Find all $𝑧 \in ℂ$ for which the columns of $(\begin{matrix} 𝑧 & 𝑖 \\ 𝑖 & 𝑧 \end{matrix})$ are linearly dependent, and for each such $𝑧$ write down a dependence relation.

Solution.

Uses. Definition 5.1.14 (The $2 \times 2$ determinant), Exercise 5.1.25, Exercise 1.1.44

Hint 1. Use

𝑖^{2} = - 1

Hint 2. By Exercise 5.1.25, the columns are dependent exactly when the determinant is

0

. Compute it as a polynomial in

𝑧

Sketch. (a)

(1 + 𝑖) (1 - 𝑖) -

2 𝑖 =

2 -

2 𝑖

. (b) The determinant is

𝑧^{2} - 𝑖^{2} = 𝑧^{2} + 1

, which vanishes exactly for

𝑧 = \pm 𝑖

. For

𝑧 = 𝑖

the columns are equal; for

𝑧 = - 𝑖

they are negatives of each other.

(a) $(1 + 𝑖) (1 - 𝑖) -$ $2 \cdot$ $𝑖 =$ $1 -$ $𝑖^{2} -$ $2 𝑖 =$ $2 -$ $2 𝑖$ .

(b) The determinant is $𝑧 \cdot 𝑧 - 𝑖 \cdot 𝑖 = 𝑧^{2} + 1$ . By Exercise 5.1.25, the columns are dependent exactly when $𝑧^{2} + 1 = 0$ , that is (Exercise 1.1.44), when $𝑧 = 𝑖$ or $𝑧 = - 𝑖$ . For $𝑧 = 𝑖$ both columns are $(𝑖, 𝑖)$ , so $𝐚_{1} - 𝐚_{2} = 𝟎$ . For $𝑧 = - 𝑖$ the columns are $(- 𝑖, 𝑖)$ and $(𝑖, - 𝑖)$ , so $𝐚_{1} + 𝐚_{2} = 𝟎$ .

Exercise 5.1.27.standard Show that

det (𝐴 𝐵) =

det 𝐴 det 𝐵

for all

𝐴, 𝐵 \in 𝐹^{2 \times 2}

Solution (via uniqueness).

Uses. Theorem 5.1.13 (The $2 \times 2$ determinant), Definition 2.4.1 (Matrix product), Definition 2.3.15 (The linear map of a matrix), Proposition 2.1.17 (Compositions of linear maps are linear)

Hint 1. Fix

𝐴

and regard

𝐷 (𝐵) = det (𝐴 𝐵)

as a function of the columns of

𝐵

Hint 2. The columns of

𝐴 𝐵

are

𝐴 𝐛_{1}

and

𝐴 𝐛_{2}

. Show that

𝐷

is multilinear and alternating, and apply part (a) of Theorem 5.1.13.

Sketch.

𝐷 (𝐛_{1}, 𝐛_{2}) =

det (𝐴 𝐛_{1}, 𝐴 𝐛_{2})

is multilinear, because

𝐱 \mapsto 𝐴 𝐱

is linear, and alternating, because equal columns of

𝐵

give equal columns of

𝐴 𝐵

. So

𝐷 (𝐵) =

det (𝐵) 𝐷 (𝐼_{2}) =

det 𝐵 det 𝐴

Fix $𝐴$ , and define $𝐷 : 𝐹^{2 \times 2} \to 𝐹$ by $𝐷 (𝐵) = det (𝐴 𝐵)$ . The columns of $𝐴 𝐵$ are $𝐴 𝐛_{1}$ and $𝐴 𝐛_{2}$ (Definition 2.4.1), so $𝐷 (𝐛_{1}, 𝐛_{2}) =$ $det (𝐴 𝐛_{1}, 𝐴 𝐛_{2})$ .

Multilinear. For fixed $𝐛_{2}$ , the map $𝐱 \mapsto$ $det (𝐴 𝐱, 𝐴 𝐛_{2})$ is the composite of the linear map $𝑇_{𝐴}$ with the linear map $𝐲 \mapsto det (𝐲, 𝐴 𝐛_{2})$ , so it is linear (Proposition 2.1.17). The same holds in the second column.

Alternating. If $𝐛_{1} = 𝐛_{2}$ , then $𝐴 𝐛_{1} = 𝐴 𝐛_{2}$ , so $𝐷 (𝐵) = 0$ .

By part (a) of Theorem 5.1.13, $𝐷 (𝐵) =$ $det (𝐵) 𝐷 (𝐼_{2}) =$ $det (𝐵) det (𝐴)$ , as desired.

Solution (by direct computation).

Uses. Definition 5.1.14 (The $2 \times 2$ determinant), Definition 2.4.1 (Matrix product)

Hint 1. Write

𝐴 = (\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix})

and

𝐵 = (\begin{matrix} 𝑝 & 𝑞 \\ 𝑟 & 𝑠 \end{matrix})

and compute

𝐴 𝐵

Hint 2. Expand

det (𝐴 𝐵)

; four of the eight products cancel in pairs.

Sketch.

𝐴 𝐵 =

(\begin{matrix} 𝑎 𝑝 + 𝑏 𝑟 & 𝑎 𝑞 + 𝑏 𝑠 \\ 𝑐 𝑝 + 𝑑 𝑟 & 𝑐 𝑞 + 𝑑 𝑠 \end{matrix})

. Expanding the determinant, the terms

𝑎 𝑐 𝑝 𝑞

and

𝑏 𝑑 𝑟 𝑠

cancel, and the rest factor as

(𝑎 𝑑 - 𝑏 𝑐) (𝑝 𝑠 - 𝑞 𝑟)

With $𝐴 = (\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix})$ and $𝐵 = (\begin{matrix} 𝑝 & 𝑞 \\ 𝑟 & 𝑠 \end{matrix})$ , we have $𝐴 𝐵 =$ $(\begin{matrix} 𝑎 𝑝 + 𝑏 𝑟 & 𝑎 𝑞 + 𝑏 𝑠 \\ 𝑐 𝑝 + 𝑑 𝑟 & 𝑐 𝑞 + 𝑑 𝑠 \end{matrix})$ , so

\begin{aligned} det (𝐴 𝐵) & = (𝑎 𝑝 + 𝑏 𝑟) (𝑐 𝑞 + 𝑑 𝑠) - (𝑎 𝑞 + 𝑏 𝑠) (𝑐 𝑝 + 𝑑 𝑟) \\ = 𝑎 𝑐 𝑝 𝑞 + 𝑎 𝑑 𝑝 𝑠 + 𝑏 𝑐 𝑞 𝑟 + 𝑏 𝑑 𝑟 𝑠 - 𝑎 𝑐 𝑝 𝑞 - 𝑎 𝑑 𝑞 𝑟 - 𝑏 𝑐 𝑝 𝑠 - 𝑏 𝑑 𝑟 𝑠 \\ = 𝑎 𝑑 (𝑝 𝑠 - 𝑞 𝑟) - 𝑏 𝑐 (𝑝 𝑠 - 𝑞 𝑟) = (𝑎 𝑑 - 𝑏 𝑐) (𝑝 𝑠 - 𝑞 𝑟) = det 𝐴 det 𝐵 . \end{aligned}

The first approach is the one that generalizes: Theorem 5.4.10 proves $det (𝐴 𝐵) =$ $det 𝐴 det 𝐵$ for all $𝑛$ , and the direct computation would be hopeless for large $𝑛$ .

Exercise 5.1.28.routine For $𝜃 \in ℝ$ , let $𝑅_{𝜃} =$ $(\begin{matrix} cos 𝜃 & - sin 𝜃 \\ sin 𝜃 & cos 𝜃 \end{matrix})$ be the rotation by $𝜃$ , and for $𝑡 \in ℝ$ let $𝑆_{𝑡} = (\begin{matrix} 1 & 𝑡 \\ 0 & 1 \end{matrix})$ . Show that

det (𝑅_{𝜃} 𝐴) = det 𝐴 = det (𝐴 𝑆_{𝑡}) for all 𝐴 \in ℝ^{2 \times 2},

and explain what each equality says about the parallelogram spanned by the columns of $𝐴$ .

Solution.

Uses. Exercise 5.1.27, Lemma 5.1.8 (Column replacement), Example 5.1.15 ( $det$ is signed area)

Hint 1. Use Exercise 5.1.27 and compute

det 𝑅_{𝜃}

and

det 𝑆_{𝑡}

Hint 2. For the interpretation: the columns of

𝑅_{𝜃} 𝐴

are the columns of

𝐴

rotated by

𝜃

, and the columns of

𝐴 𝑆_{𝑡}

are

𝐚_{1}

and

𝐚_{2} + 𝑡 𝐚_{1}

Sketch.

det 𝑅_{𝜃} =

{cos}^{2} 𝜃 +

{sin}^{2} 𝜃 =

1

and

det 𝑆_{𝑡} = 1

, so both equalities follow from multiplicativity. Rotating a parallelogram, or sliding one edge parallel to the other, preserves its signed area.

By Exercise 5.1.27, $det (𝑅_{𝜃} 𝐴) =$ $det (𝑅_{𝜃}) det 𝐴$ and $det (𝐴 𝑆_{𝑡}) =$ $det 𝐴 det 𝑆_{𝑡}$ . Since $det 𝑅_{𝜃} =$ ${cos}^{2} 𝜃 +$ ${sin}^{2} 𝜃 =$ $1$ and $det 𝑆_{𝑡} =$ $1 \cdot$ $1 -$ $𝑡 \cdot$ $0 =$ $1$ , both equalities hold. (The second also follows from Lemma 5.1.8, since $𝐴 𝑆_{𝑡}$ has columns $𝐚_{1}$ and $𝐚_{2} + 𝑡 𝐚_{1}$ .)

The columns of $𝑅_{𝜃} 𝐴$ are $𝑅_{𝜃} 𝐚_{1}$ and $𝑅_{𝜃} 𝐚_{2}$ , so $𝑃 (𝑅_{𝜃} 𝐚_{1}, 𝑅_{𝜃} 𝐚_{2})$ is $𝑃 (𝐚_{1}, 𝐚_{2})$ rotated by $𝜃$ , and the first equality says that rotation preserves signed area (area and orientation). The second says that sliding the edge $𝐚_{2}$ parallel to the edge $𝐚_{1}$ preserves signed area. These are the two facts used informally in Intuition (Signed area is linear in each column); with Example 5.1.15 they are now theorems.

Exercise 5.1.29.standard Find all multilinear alternating functions

𝐷 : 𝐹^{2 \times 2} \to 𝐹

with

𝐷 (\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}) = 6

when (a)

𝐹 = ℚ

; (b)

𝐹 = 𝔽_{2}

; (c)

𝐹 = 𝔽_{3}

Solution.

Uses. Theorem 5.1.13 (The $2 \times 2$ determinant), Definition 5.1.14 (The $2 \times 2$ determinant)

Hint 1. By Theorem 5.1.13, every multilinear alternating

𝐷

𝑐 det

for

𝑐 = 𝐷 (𝐼_{2})

. Conversely, which multiples of

det

are multilinear and alternating?

Hint 2. So the condition is

𝑐 det (\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}) = 6

, that is,

- 2 𝑐 = 6

. Solve it in each field.

Sketch. The candidates are exactly

𝐷 = 𝑐 det

with

𝑐 \in 𝐹

, and the condition is

- 2 𝑐 = 6

. Over

ℚ

𝑐 = - 3

. Over

𝔽_{2}

, it reads

0 = 0

, so

𝐷 = 0

and

𝐷 = det

both work. Over

𝔽_{3}

, it reads

𝑐 = 0

, so only

𝐷 = 0

By part (a) of Theorem 5.1.13, every multilinear alternating $𝐷$ equals $𝑐 det$ with $𝑐 = 𝐷 (𝐼_{2})$ . Conversely, for every $𝑐 \in 𝐹$ the function $𝑐 det$ is multilinear and alternating, since multiplying the identities defining these properties by $𝑐$ preserves them. So the functions in question are the $𝑐 det$ with $𝑐 det (\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}) = 6$ . Since $det (\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}) = 4 - 6 = - 2$ , the condition is $- 2 𝑐 = 6$ .

(a) Over $ℚ$ , $𝑐 = - 3$ , so $𝐷 = - 3 det$ is the only one.

(b) Over $𝔽_{2}$ , $- 2 = 0$ and $6 = 0$ , so the condition holds for both $𝑐 = 0$ and $𝑐 = 1$ : the answers are $𝐷 = 0$ and $𝐷 = det$ .

(c) Over $𝔽_{3}$ , $- 2 = 1$ and $6 = 0$ , so $𝑐 = 0$ , and $𝐷 = 0$ is the only one.

Exercise 5.1.30.standard For which fields

𝐹

is the function

𝐴 \mapsto {(det 𝐴)}^{2}

a determinant function on

𝐹^{2 \times 2}

Solution.

Uses. Definition 5.1.1 (Determinant function), Theorem 5.1.13 (The $2 \times 2$ determinant), Exercise 1.1.18

Hint 1. Normalization and alternation hold over every field. Test multilinearity on

(𝑐 𝐞_{1}, 𝐞_{2})

Hint 2. Multilinearity would force

𝑐^{2} = 𝑐

for every

𝑐 \in 𝐹

. Which fields satisfy this? (See Exercise 1.1.18.)

Sketch.

{(det 𝐼_{2})}^{2} = 1

, and equal columns give

0^{2} = 0

. Linearity in the first column forces

𝑐^{2} =

{(det (𝑐 𝐞_{1}, 𝐞_{2}))}^{2} =

𝑐

for all

𝑐

, so

𝐹 = {0, 1}

, which is

𝔽_{2}

. Over

𝔽_{2}

𝑥^{2} = 𝑥

for every

𝑥

, so

{(det 𝐴)}^{2} = det 𝐴

Let $𝐸 (𝐴) = {(det 𝐴)}^{2}$ . Then $𝐸 (𝐼_{2}) = 1$ , and if the columns of $𝐴$ are equal, then $det 𝐴 = 0$ and $𝐸 (𝐴) = 0$ . So $𝐸$ is a determinant function if and only if it is multilinear.

Suppose $𝐸$ is multilinear. For every $𝑐 \in 𝐹$ , linearity in the first column gives $𝐸 (𝑐 𝐞_{1}, 𝐞_{2}) =$ $𝑐 𝐸 (𝐞_{1}, 𝐞_{2}) =$ $𝑐$ , while directly $𝐸 (𝑐 𝐞_{1}, 𝐞_{2}) =$ ${(det (\begin{matrix} 𝑐 & 0 \\ 0 & 1 \end{matrix}))}^{2} =$ $𝑐^{2}$ . So $𝑐^{2} = 𝑐$ for every $𝑐 \in 𝐹$ , and by Exercise 1.1.18 every element of $𝐹$ is $0$ or $1$ . Thus $𝐹 = {0, 1}$ . In such a field $1 + 1 \neq 1$ (else $1 = 0$ ), so $1 + 1 = 0$ and $𝐹$ has the tables of $𝔽_{2}$ .

Conversely, in $𝔽_{2}$ we have $0^{2} = 0$ and $1^{2} = 1$ , so $𝐸 (𝐴) = det 𝐴$ for every $𝐴$ , and $𝐸$ is a determinant function by Theorem 5.1.13. So the answer is: exactly when $𝐹 = 𝔽_{2}$ .

Exercise 5.1.31.standard Prove or disprove each statement.

(a)If $𝐷 : 𝐹^{2 \times 2} \to 𝐹$ is multilinear and alternating and $𝐷 (𝐴) = 0$ for some invertible $𝐴$ , then $𝐷 = 0$ .
(b)If $𝐷 : ℝ^{2 \times 2} \to ℝ$ is alternating and $𝐷 (𝑐 𝐴) =$ $𝑐^{2} 𝐷 (𝐴)$ for all $𝑐 \in ℝ$ and all $𝐴$ , then $𝐷$ is a scalar multiple of $det$ .
(c)If $𝐷 : 𝐹^{3 \times 3} \to 𝐹$ is multilinear and alternating, then $𝐷 (𝐚_{1}, 𝐚_{2}, 𝐚_{3}) =$ $𝐷 (𝐚_{2}, 𝐚_{3}, 𝐚_{1})$ for all $𝐚_{1}, 𝐚_{2}, 𝐚_{3} \in 𝐹^{3}$ .

Solution.

Uses. Theorem 5.1.13 (The $2 \times 2$ determinant), Exercise 5.1.25, Proposition 5.1.10 (Alternating implies antisymmetric), Non-example 5.1.3 (One axiom at a time)

Hint 1. (a) Use part (a) of Theorem 5.1.13 and Exercise 5.1.25.

Hint 2. (b) The hypothesis

𝐷 (𝑐 𝐴) =

𝑐^{2} 𝐷 (𝐴)

is weaker than multilinearity. Look back at Non-example 5.1.3.

Hint 3. (c) Count the interchanges needed to sort

𝐚_{2}, 𝐚_{3}, 𝐚_{1}

Sketch. (a) True:

0 =

𝐷 (𝐴) =

det (𝐴) 𝐷 (𝐼_{2})

with

det 𝐴 \neq 0

, so

𝐷 (𝐼_{2}) = 0

and

𝐷 = 0

. (b) False:

𝐷 = | det |

. (c) True: two interchanges.

(a) True. By part (a) of Theorem 5.1.13, $𝐷 = 𝐷 (𝐼_{2}) det$ . Since $𝐴$ is invertible, $det 𝐴 \neq 0$ by Exercise 5.1.25, so $0 =$ $𝐷 (𝐴) =$ $𝐷 (𝐼_{2}) det 𝐴$ forces $𝐷 (𝐼_{2}) = 0$ , and then $𝐷 = 0$ .

(b) False. Let $𝐷 (𝐴) = | det 𝐴 |$ . It is alternating (Non-example 5.1.3), and $𝐷 (𝑐 𝐴) =$ $| 𝑐^{2} det 𝐴 | =$ $𝑐^{2} 𝐷 (𝐴)$ . If $𝐷 = 𝜆 det$ for some $𝜆 \in ℝ$ , then evaluating at $𝐼_{2}$ gives $𝜆 = 1$ , and evaluating at $(\begin{matrix} - 1 & 0 \\ 0 & 1 \end{matrix})$ gives $1 = - 𝜆$ . These are incompatible.

(c) True. Interchanging columns $1$ and $3$ and then columns $2$ and $3$ , Proposition 5.1.10 gives

𝐷 (𝐚_{2}, 𝐚_{3}, 𝐚_{1}) = - 𝐷 (𝐚_{1}, 𝐚_{3}, 𝐚_{2}) = 𝐷 (𝐚_{1}, 𝐚_{2}, 𝐚_{3}) .

Exercise 5.1.32.standard For $𝐱, 𝐲 \in 𝐹^{3}$ , define the cross product

𝐱 \times 𝐲 = (𝑥_{2} 𝑦_{3} - 𝑥_{3} 𝑦_{2}, 𝑥_{3} 𝑦_{1} - 𝑥_{1} 𝑦_{3}, 𝑥_{1} 𝑦_{2} - 𝑥_{2} 𝑦_{1}),

and define $𝐷 : 𝐹^{3 \times 3} \to 𝐹$ by

𝐷 (𝐱, 𝐲, 𝐳) = {(𝐱 \times 𝐲)}_{1} 𝑧_{1} + {(𝐱 \times 𝐲)}_{2} 𝑧_{2} + {(𝐱 \times 𝐲)}_{3} 𝑧_{3} .

Show that $𝐷$ is a determinant function on $𝐹^{3 \times 3}$ . (Over $ℝ$ this is the scalar triple product. So determinant functions exist for $𝑛 = 3$ ; Section 5.3 treats all $𝑛$ .)

Solution.

Uses. Definition 5.1.1 (Determinant function), Theorem 5.1.13 (The $2 \times 2$ determinant)

Hint 1. Each coordinate of

𝐱 \times 𝐲

is a

2 \times 2

determinant built from two coordinates of

𝐱

and of

𝐲

. Use this for multilinearity.

Hint 2. For alternation there are three cases:

𝐱 = 𝐲

𝐲 = 𝐳

, and

𝐱 = 𝐳

. In the last two, expand and watch the six terms cancel in pairs.

Sketch. Each

{(𝐱 \times 𝐲)}_{𝑘}

is bilinear in

𝐱, 𝐲

, so each term

{(𝐱 \times 𝐲)}_{𝑘} 𝑧_{𝑘}

is linear in each of

𝐱, 𝐲, 𝐳

𝐱 \times 𝐱 = 𝟎

, and

(𝐱 \times 𝐲) \cdot 𝐲

and

(𝐱 \times 𝐲) \cdot 𝐱

expand to six terms cancelling in pairs.

𝐞_{1} \times 𝐞_{2} = 𝐞_{3}

Multilinear. The coordinate ${(𝐱 \times 𝐲)}_{3} =$ $𝑥_{1} 𝑦_{2} -$ $𝑥_{2} 𝑦_{1}$ is the determinant of $(\begin{matrix} 𝑥_{1} & 𝑦_{1} \\ 𝑥_{2} & 𝑦_{2} \end{matrix})$ , and similarly for the other two coordinates, so by Theorem 5.1.13 each coordinate of $𝐱 \times 𝐲$ is linear in $𝐱$ for fixed $𝐲$ , and linear in $𝐲$ for fixed $𝐱$ . Hence each term ${(𝐱 \times 𝐲)}_{𝑘} 𝑧_{𝑘}$ is linear in each of $𝐱$ , $𝐲$ , and $𝐳$ when the other two are fixed, and so is their sum $𝐷$ .

Alternating. If $𝐱 = 𝐲$ , every coordinate of $𝐱 \times 𝐱$ has the form $𝑥_{𝑖} 𝑥_{𝑗} -$ $𝑥_{𝑗} 𝑥_{𝑖} =$ $0$ , so $𝐷 = 0$ . If $𝐲 = 𝐳$ ,

𝐷 = (𝑥_{2} 𝑦_{3} - 𝑥_{3} 𝑦_{2}) 𝑦_{1} + (𝑥_{3} 𝑦_{1} - 𝑥_{1} 𝑦_{3}) 𝑦_{2} + (𝑥_{1} 𝑦_{2} - 𝑥_{2} 𝑦_{1}) 𝑦_{3},

and the six terms cancel in pairs: $𝑥_{2} 𝑦_{1} 𝑦_{3}$ with $- 𝑥_{2} 𝑦_{1} 𝑦_{3}$ , $- 𝑥_{3} 𝑦_{1} 𝑦_{2}$ with $𝑥_{3} 𝑦_{1} 𝑦_{2}$ , and $- 𝑥_{1} 𝑦_{2} 𝑦_{3}$ with $𝑥_{1} 𝑦_{2} 𝑦_{3}$ . If $𝐱 = 𝐳$ ,

𝐷 = (𝑥_{2} 𝑦_{3} - 𝑥_{3} 𝑦_{2}) 𝑥_{1} + (𝑥_{3} 𝑦_{1} - 𝑥_{1} 𝑦_{3}) 𝑥_{2} + (𝑥_{1} 𝑦_{2} - 𝑥_{2} 𝑦_{1}) 𝑥_{3},

and again $𝑥_{1} 𝑥_{2} 𝑦_{3}$ , $𝑥_{1} 𝑥_{3} 𝑦_{2}$ , and $𝑥_{2} 𝑥_{3} 𝑦_{1}$ each appear once with each sign.

Normalized. $𝐞_{1} \times$ $𝐞_{2} =$ $(0 \cdot 0 - 0 \cdot 1, 0 \cdot 0 - 1 \cdot 0, 1 \cdot 1 - 0 \cdot 0) =$ $𝐞_{3}$ , so $𝐷 (𝐼_{3}) =$ $𝐷 (𝐞_{1}, 𝐞_{2}, 𝐞_{3}) =$ $1$ .

Exercise 5.1.33.standard Let $𝐴 = (\begin{matrix} 1 & 0 & 2 \\ 3 & 1 & 0 \\ 0 & 4 & 1 \end{matrix})$ , and let $𝐷 : 𝐹^{3 \times 3} \to 𝐹$ be multilinear and alternating.

(a)Show that $𝐷 (𝐴) = 25 𝐷 (𝐼_{3})$ by expanding each column in the standard basis.
(b)Regard $𝐴$ as a matrix over $𝔽_{5}$ . Find a nontrivial linear dependence among its columns, and explain why this is consistent with (a).
(c)Regard $𝐴$ as a matrix over $ℚ$ . Show that $𝐴$ is invertible. (A determinant function on $ℚ^{3 \times 3}$ exists by Exercise 5.1.32.)

Solution.

Uses. Proposition 5.1.10 (Alternating implies antisymmetric), Proposition 5.1.17 (Dependent columns), Corollary 5.1.18 (Non-invertible matrices), Exercise 5.1.32

Techniques. Expand by multilinearity (Technique 5.1.7)

Hint 1. The columns are

𝐞_{1} + 3 𝐞_{2}

𝐞_{2} + 4 𝐞_{3}

, and

2 𝐞_{1} + 𝐞_{3}

. Expanding gives eight terms. Discard those with a repeated basis vector.

Hint 2. Exactly two terms survive. One of them needs

𝐷 (𝐞_{2}, 𝐞_{3}, 𝐞_{1})

; compute it with interchanges.

Hint 3. For (b), solve

𝐴 𝐱 = 𝟎

over

𝔽_{5}

, starting from the first row.

Sketch. The surviving terms are

𝐷 (𝐞_{1}, 𝐞_{2}, 𝐞_{3})

and

3 \cdot

4 \cdot

2 \cdot

𝐷 (𝐞_{2}, 𝐞_{3}, 𝐞_{1}) =

24 𝐷 (𝐼_{3})

, so

𝐷 (𝐴) = 25 𝐷 (𝐼_{3})

. Over

𝔽_{5}

3 𝐚_{1} +

𝐚_{2} +

𝐚_{3} =

(5, 10, 5) =

𝟎

, and

25 = 0

. Over

ℚ

, a determinant function gives

𝐷 (𝐴) = 25 \neq 0

(a) The columns are $𝐚_{1} = 𝐞_{1} + 3 𝐞_{2}$ , $𝐚_{2} = 𝐞_{2} + 4 𝐞_{3}$ , and $𝐚_{3} = 2 𝐞_{1} + 𝐞_{3}$ . Expanding each column by multilinearity writes $𝐷 (𝐴)$ as a sum of eight terms, one for each choice of a basis vector from each column. A term in which some $𝐞_{𝑘}$ is chosen twice is $0$ because $𝐷$ is alternating. If $𝐞_{1}$ is chosen from $𝐚_{1}$ , then $𝐚_{3}$ must contribute $𝐞_{3}$ , so $𝐚_{2}$ must contribute $𝐞_{2}$ . If $𝐞_{2}$ is chosen from $𝐚_{1}$ , then $𝐚_{2}$ must contribute $𝐞_{3}$ , so $𝐚_{3}$ must contribute $𝐞_{1}$ . So only two terms survive:

𝐷 (𝐴) = 1 \cdot 1 \cdot 1 \cdot 𝐷 (𝐞_{1}, 𝐞_{2}, 𝐞_{3}) + 3 \cdot 4 \cdot 2 \cdot 𝐷 (𝐞_{2}, 𝐞_{3}, 𝐞_{1}) .

By two interchanges (Proposition 5.1.10), $𝐷 (𝐞_{2}, 𝐞_{3}, 𝐞_{1}) =$ $- 𝐷 (𝐞_{1}, 𝐞_{3}, 𝐞_{2}) =$ $𝐷 (𝐞_{1}, 𝐞_{2}, 𝐞_{3})$ . Hence $𝐷 (𝐴) =$ $(1 + 24) 𝐷 (𝐼_{3}) =$ $25 𝐷 (𝐼_{3})$ .

(b) Over $𝔽_{5}$ , solve $𝐴 𝐱 = 𝟎$ : the first row gives $𝑥_{1} = - 2 𝑥_{3} = 3 𝑥_{3}$ , the second gives $𝑥_{2} =$ $- 3 𝑥_{1} =$ $- 9 𝑥_{3} =$ $𝑥_{3}$ , and the third, $4 𝑥_{2} + 𝑥_{3} = 5 𝑥_{3} = 0$ , then holds automatically. Taking $𝑥_{3} = 1$ ,

3 𝐚_{1} + 𝐚_{2} + 𝐚_{3} = (3 + 0 + 2, 9 + 1 + 0, 0 + 4 + 1) = (5, 10, 5) = 𝟎 in 𝔽_{5}^{3} .

So the columns are dependent over $𝔽_{5}$ , and Proposition 5.1.17 gives $𝐷 (𝐴) = 0$ . This agrees with (a), because $25 = 0$ in $𝔽_{5}$ .

(c) Let $𝐷$ be a determinant function on $ℚ^{3 \times 3}$ . By (a), $𝐷 (𝐴) = 25 \neq 0$ , so $𝐴$ is invertible by Corollary 5.1.18.

Exercise 5.1.34.standard Let ${Alt}_{𝑛}$ be the set of multilinear alternating functions $𝐹^{𝑛 \times 𝑛} \to 𝐹$ .

(a)Show that ${Alt}_{𝑛}$ is a subspace of the vector space of all functions $𝐹^{𝑛 \times 𝑛} \to 𝐹$ .
(b)Show that ${Alt}_{2} = span (det)$ , and that $dim {Alt}_{2} = 1$ .

Solution.

Uses. Lemma 2.1.22 (Functions into a vector space), Definition 1.4.1 (Subspace), Theorem 1.4.3 (Subspace criterion), Proposition 2.1.23 ( $L (𝑉, 𝑊)$ is a vector space), Theorem 5.1.13 (The $2 \times 2$ determinant), Definition 4.1.5 (Dimension)

Hint 1. For (a), check that

0 \in {Alt}_{𝑛}

and that

𝐷 + 𝑠 𝐷^{'}

is multilinear and alternating whenever

𝐷

and

𝐷^{'}

are.

Hint 2. For (b), part (a) of Theorem 5.1.13 says every element of

{Alt}_{2}

is a multiple of

det

Sketch. Sums and scalar multiples of linear maps are linear, and

0 + 0 = 0

, so

{Alt}_{𝑛}

is closed under linear combinations. By Theorem 5.1.13,

{Alt}_{2}

consists exactly of the multiples of

det \neq 0

, so

det

alone is a basis.

(a) The zero function is multilinear and alternating. Let $𝐷, 𝐷^{'} \in {Alt}_{𝑛}$ and $𝑠 \in 𝐹$ , and let $𝐸 = 𝐷 + 𝑠 𝐷^{'}$ . Fix a column index $𝑗$ and the other columns. The map $𝐱 \mapsto 𝐸 (\dots, 𝐱, \dots)$ is $𝐱 \mapsto$ $𝐷 (\dots, 𝐱, \dots) +$ $𝑠 𝐷^{'} (\dots, 𝐱, \dots)$ , a linear combination of two linear maps $𝐹^{𝑛} \to 𝐹$ , hence linear (Proposition 2.1.23). If $𝐴$ has two equal columns, then $𝐸 (𝐴) =$ $𝐷 (𝐴) +$ $𝑠 𝐷^{'} (𝐴) =$ $0 +$ $𝑠 \cdot$ $0 =$ $0$ . So $𝐸 \in {Alt}_{𝑛}$ , and ${Alt}_{𝑛}$ is a subspace by Theorem 1.4.3.

(b) By part (b) of Theorem 5.1.13, $det \in {Alt}_{2}$ , so $span (det) \subseteq {Alt}_{2}$ by (a). By part (a) of Theorem 5.1.13, every $𝐷 \in {Alt}_{2}$ equals $𝐷 (𝐼_{2}) det \in$ $span (det)$ . Hence ${Alt}_{2} = span (det)$ . Since $det (𝐼_{2}) = 1$ , $det$ is not the zero function, so the one-element list $det$ is linearly independent and is a basis of ${Alt}_{2}$ . Thus $dim {Alt}_{2} = 1$ .

The expansion in Intuition (What the proof did, and where it leads) suggests that $dim {Alt}_{𝑛} = 1$ for every $𝑛$ ; this follows from Section 5.3.

Exercise 5.1.35.standard Let

𝐷 : 𝐹^{𝑛 \times 𝑛} \to 𝐹

be multilinear, and suppose that

𝐷 (𝐴) = 0

whenever two adjacent columns of

𝐴

are equal, that is, whenever

𝐚_{𝑖} = 𝐚_{𝑖 + 1}

for some

𝑖

. Show that

𝐷

is alternating.

Solution.

Uses. Proposition 5.1.10 (Alternating implies antisymmetric), Definition 5.1.1 (Determinant function)

Hint 1. First show that interchanging two adjacent columns negates

𝐷

. Which hypothesis did the first proof of Proposition 5.1.10 actually use?

Hint 2. If

𝐚_{𝑖} = 𝐚_{𝑗}

with

𝑗 > 𝑖 + 1

, move column

𝑗

to position

𝑖 + 1

by interchanging adjacent columns.

Sketch. The expansion proof of Proposition 5.1.10, with

𝑗 = 𝑖 + 1

, uses alternation only for columns

𝑖

and

𝑖 + 1

, so adjacent interchanges negate

𝐷

. If

𝐚_{𝑖} = 𝐚_{𝑗}

with

𝑖 < 𝑗

, then

𝑗 - 𝑖 - 1

adjacent interchanges bring column

𝑗

next to column

𝑖

, giving a matrix

𝐴^{'}

with

𝐷 (𝐴^{'}) = 0

and

𝐷 (𝐴) = \pm 𝐷 (𝐴^{'})

Step 1. Fix $𝑖$ and the columns other than $𝑖$ and $𝑖 + 1$ , and let $𝑓 (𝐱, 𝐲)$ be the value of $𝐷$ with $𝐱$ in column $𝑖$ and $𝐲$ in column $𝑖 + 1$ . By hypothesis $𝑓 (𝐱, 𝐱) = 0$ for every $𝐱$ . The first proof of Proposition 5.1.10 used nothing else besides multilinearity, so it gives $𝑓 (𝐲, 𝐱) =$ $- 𝑓 (𝐱, 𝐲)$ . That is, interchanging two adjacent columns negates $𝐷$ .

Step 2. Let $𝐴$ have $𝐚_{𝑖} = 𝐚_{𝑗}$ with $𝑖 < 𝑗$ . If $𝑗 = 𝑖 + 1$ , then $𝐷 (𝐴) = 0$ by hypothesis. Otherwise, interchange columns $𝑗 - 1$ and $𝑗$ , then columns $𝑗 - 2$ and $𝑗 - 1$ , and so on, ending with columns $𝑖 + 1$ and $𝑖 + 2$ . These $𝑗 - 𝑖 - 1$ adjacent interchanges move the vector $𝐚_{𝑗}$ from position $𝑗$ to position $𝑖 + 1$ , and never touch column $𝑖$ . The resulting matrix $𝐴^{'}$ has $𝐚_{𝑖}$ in column $𝑖$ and $𝐚_{𝑗} = 𝐚_{𝑖}$ in column $𝑖 + 1$ , so $𝐷 (𝐴^{'}) = 0$ . By Step 1, $𝐷 (𝐴) =$ ${(- 1)}^{𝑗 - 𝑖 - 1} 𝐷 (𝐴^{'}) =$ $0$ . Hence $𝐷$ is alternating.

Exercise 5.1.36.standard Let

𝑘 > 𝑛

, and let

𝑓 : 𝐹^{𝑛} \times \dots \times 𝐹^{𝑛} \to 𝐹

(with

𝑘

factors) be multilinear and alternating, in the sense of Remark 5.1.6. Show that

𝑓 = 0

. Interpret the result geometrically for

𝑘 = 3

and

𝑛 = 2

Solution.

Uses. Proposition 5.1.17 (Dependent columns), Theorem 4.1.10 (Independent lists are short), Remark 5.1.6

Hint 1. What can you say about any

𝑘

vectors in

𝐹^{𝑛}

when

𝑘 > 𝑛

Hint 2. They are dependent. Now repeat the first proof of Proposition 5.1.17 for

𝑓

Sketch. Any

𝑘 > 𝑛 = dim 𝐹^{𝑛}

vectors are dependent, so one of them is a combination of the others. Expanding

𝑓

in that slot gives terms with a repeated argument.

Let $𝐯_{1}, \dots, 𝐯_{𝑘} \in 𝐹^{𝑛}$ . Since $dim 𝐹^{𝑛} = 𝑛 < 𝑘$ , the list $𝐯_{1}, \dots, 𝐯_{𝑘}$ is linearly dependent (Theorem 4.1.10). As in the first proof of Proposition 5.1.17, some $𝐯_{𝑚}$ equals $\sum_{𝑗 \neq 𝑚} 𝑏_{𝑗} 𝐯_{𝑗}$ , and linearity in slot $𝑚$ gives

𝑓 (𝐯_{1}, \dots, 𝐯_{𝑘}) = \sum_{𝑗 \neq 𝑚} 𝑏_{𝑗} 𝑓 (𝐯_{1}, \dots, 𝐯_{𝑗}, \dots, 𝐯_{𝑘}),

where in the $𝑗$ -th term $𝐯_{𝑗}$ occupies both slot $𝑗$ and slot $𝑚$ . Each term is $0$ because $𝑓$ is alternating. So $𝑓 (𝐯_{1}, \dots, 𝐯_{𝑘}) = 0$ for all inputs, that is, $𝑓 = 0$ .

For $𝑘 = 3$ and $𝑛 = 2$ , $𝑓$ would measure a signed volume of the parallelepiped spanned by three vectors in the plane. Such a parallelepiped is flat, so its volume is $0$ , and the result says that no alternating multilinear function can do better.

Exercise 5.1.37.standard For $𝐱, 𝐲 \in 𝐹^{3}$ and $1 \leq 𝑖 < 𝑗 \leq 3$ , let $𝑓_{𝑖 𝑗} (𝐱, 𝐲) =$ $𝑥_{𝑖} 𝑦_{𝑗} -$ $𝑥_{𝑗} 𝑦_{𝑖}$ .

(a)Show that $𝑓_{12}$ , $𝑓_{13}$ , and $𝑓_{23}$ are multilinear and alternating functions of two vectors in $𝐹^{3}$ .
(b)Show that they are linearly independent, as elements of the vector space of functions $𝐹^{3} \times 𝐹^{3} \to 𝐹$ .
(c)Show that every multilinear alternating $𝑓 : 𝐹^{3} \times 𝐹^{3} \to 𝐹$ satisfies $𝑓 =$ $\sum_{𝑖 < 𝑗} 𝑓 (𝐞_{𝑖}, 𝐞_{𝑗}) 𝑓_{𝑖 𝑗}$ .

So for two vectors in $𝐹^{3}$ there is a three-dimensional space of “signed areas”, not a single one up to scaling. This is one reason determinants are only defined for square matrices.

Solution.

Uses. Theorem 5.1.13 (The $2 \times 2$ determinant), Proposition 5.1.10 (Alternating implies antisymmetric), Remark 5.1.6, Definition 1.5.1 (Linear independence)

Techniques. Expand by multilinearity (Technique 5.1.7)

Hint 1.

𝑓_{𝑖 𝑗} (𝐱, 𝐲)

is the

2 \times 2

determinant of rows

𝑖

and

𝑗

of the

3 \times 2

matrix with columns

𝐱, 𝐲

Hint 2. For (b), evaluate a vanishing combination at

(𝐞_{1}, 𝐞_{2})

(𝐞_{1}, 𝐞_{3})

, and

(𝐞_{2}, 𝐞_{3})

Hint 3. For (c), write

𝐱 = \sum_{𝑖} 𝑥_{𝑖} 𝐞_{𝑖}

and

𝐲 = \sum_{𝑗} 𝑦_{𝑗} 𝐞_{𝑗}

and expand. Group the terms

(𝑖, 𝑗)

and

(𝑗, 𝑖)

Sketch. (a) Each

𝑓_{𝑖 𝑗}

is a

2 \times 2

determinant of coordinates. (b)

𝑓_{𝑖 𝑗} (𝐞_{𝑘}, 𝐞_{𝑙})

1

(𝑘, 𝑙) = (𝑖, 𝑗)

and

0

for the other pairs

𝑘 < 𝑙

. (c) Expanding gives

𝑓 (𝐱, 𝐲) =

\sum_{𝑖, 𝑗} 𝑥_{𝑖} 𝑦_{𝑗} 𝑓 (𝐞_{𝑖}, 𝐞_{𝑗})

; diagonal terms vanish, and antisymmetry pairs

(𝑖, 𝑗)

with

(𝑗, 𝑖)

into

(𝑥_{𝑖} 𝑦_{𝑗} - 𝑥_{𝑗} 𝑦_{𝑖}) 𝑓 (𝐞_{𝑖}, 𝐞_{𝑗})

(a) $𝑓_{𝑖 𝑗} (𝐱, 𝐲) =$ $det (\begin{matrix} 𝑥_{𝑖} & 𝑦_{𝑖} \\ 𝑥_{𝑗} & 𝑦_{𝑗} \end{matrix})$ , and the coordinates $𝑥_{𝑖}, 𝑥_{𝑗}$ depend linearly on $𝐱$ . By Theorem 5.1.13, $𝑓_{𝑖 𝑗}$ is linear in $𝐱$ for fixed $𝐲$ and in $𝐲$ for fixed $𝐱$ , and $𝑓_{𝑖 𝑗} (𝐱, 𝐱) =$ $𝑥_{𝑖} 𝑥_{𝑗} -$ $𝑥_{𝑗} 𝑥_{𝑖} =$ $0$ .

(b) For $𝑘 < 𝑙$ , $𝑓_{𝑖 𝑗} (𝐞_{𝑘}, 𝐞_{𝑙}) =$ $1$ if $(𝑘, 𝑙) = (𝑖, 𝑗)$ , and $0$ otherwise. If $𝑐_{12} 𝑓_{12} +$ $𝑐_{13} 𝑓_{13} +$ $𝑐_{23} 𝑓_{23} =$ $0$ , evaluating at $(𝐞_{1}, 𝐞_{2})$ , $(𝐞_{1}, 𝐞_{3})$ , and $(𝐞_{2}, 𝐞_{3})$ gives $𝑐_{12} = 𝑐_{13} = 𝑐_{23} = 0$ .

(c) Expanding $𝐱 = \sum_{𝑖} 𝑥_{𝑖} 𝐞_{𝑖}$ and $𝐲 = \sum_{𝑗} 𝑦_{𝑗} 𝐞_{𝑗}$ ,

𝑓 (𝐱, 𝐲) = \sum_{𝑖 = 1}^{3} \sum_{𝑗 = 1}^{3} 𝑥_{𝑖} 𝑦_{𝑗} 𝑓 (𝐞_{𝑖}, 𝐞_{𝑗}) .

The terms with $𝑖 = 𝑗$ vanish since $𝑓$ is alternating. For $𝑖 < 𝑗$ , $𝑓 (𝐞_{𝑗}, 𝐞_{𝑖}) =$ $- 𝑓 (𝐞_{𝑖}, 𝐞_{𝑗})$ by Proposition 5.1.10 (which holds for $𝑓$ by Remark 5.1.6), so the terms $(𝑖, 𝑗)$ and $(𝑗, 𝑖)$ combine to $(𝑥_{𝑖} 𝑦_{𝑗} - 𝑥_{𝑗} 𝑦_{𝑖}) 𝑓 (𝐞_{𝑖}, 𝐞_{𝑗})$ . Hence

𝑓 (𝐱, 𝐲) = \sum_{𝑖 < 𝑗} 𝑓 (𝐞_{𝑖}, 𝐞_{𝑗}) 𝑓_{𝑖 𝑗} (𝐱, 𝐲),

as desired. With (a) and (b), the $𝑓_{𝑖 𝑗}$ form a basis of the space of multilinear alternating functions of two vectors in $𝐹^{3}$ , which therefore has dimension $3$ , compared with dimension $1$ for two vectors in $𝐹^{2}$ (Exercise 5.1.34).

Exercise 5.1.38.standard

(a)Show that $det (𝐴 + 𝑡 𝐼_{2}) =$ $𝑡^{2} +$ $(tr 𝐴) 𝑡 +$ $det 𝐴$ for all $𝐴 \in 𝐹^{2 \times 2}$ and $𝑡 \in 𝐹$ .
(b)Let $𝐴 = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})$ , the rotation by $𝜋 / 2$ . Find all $𝑡 \in ℂ$ for which $𝐴 + 𝑡 𝐼_{2}$ is not invertible over $ℂ$ , and all $𝑡 \in ℝ$ for which it is not invertible over $ℝ$ .

Solution.

Uses. Theorem 5.1.13 (The $2 \times 2$ determinant), Definition 2.4.26 (Trace), Exercise 5.1.25, Exercise 1.1.44

Techniques. Expand by multilinearity (Technique 5.1.7)

Hint 1. The columns of

𝐴 + 𝑡 𝐼_{2}

are

𝐚_{1} + 𝑡 𝐞_{1}

and

𝐚_{2} + 𝑡 𝐞_{2}

. Expand into four terms.

Hint 2. The mixed terms are

𝑡 det (𝐚_{1}, 𝐞_{2})

and

𝑡 det (𝐞_{1}, 𝐚_{2})

. Compute them from the formula.

Hint 3. For (b),

𝐴 + 𝑡 𝐼_{2}

is invertible exactly when its determinant is nonzero (Exercise 5.1.25).

Sketch. Expanding,

det (𝐴 + 𝑡 𝐼_{2}) =

det 𝐴 +

𝑡 det (𝐚_{1}, 𝐞_{2}) +

𝑡 det (𝐞_{1}, 𝐚_{2}) +

𝑡^{2}

, and the middle terms are

𝑡 𝑎

and

𝑡 𝑑

. For the rotation,

det (𝐴 + 𝑡 𝐼_{2}) =

𝑡^{2} +

1

, which vanishes for

𝑡 = \pm 𝑖

and for no real

𝑡

(a) Let $𝐴 = (\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix})$ . The columns of $𝐴 + 𝑡 𝐼_{2}$ are $𝐚_{1} + 𝑡 𝐞_{1}$ and $𝐚_{2} + 𝑡 𝐞_{2}$ . Expanding by multilinearity,

det (𝐴 + 𝑡 𝐼_{2}) = det (𝐚_{1}, 𝐚_{2}) + 𝑡 det (𝐚_{1}, 𝐞_{2}) + 𝑡 det (𝐞_{1}, 𝐚_{2}) + 𝑡^{2} det (𝐞_{1}, 𝐞_{2}) .

Here $det (𝐚_{1}, 𝐞_{2}) =$ $det (\begin{matrix} 𝑎 & 0 \\ 𝑐 & 1 \end{matrix}) =$ $𝑎$ , $det (𝐞_{1}, 𝐚_{2}) =$ $det (\begin{matrix} 1 & 𝑏 \\ 0 & 𝑑 \end{matrix}) =$ $𝑑$ , and $det (𝐞_{1}, 𝐞_{2}) = 1$ . So $det (𝐴 + 𝑡 𝐼_{2}) =$ $det 𝐴 +$ $(𝑎 + 𝑑) 𝑡 +$ $𝑡^{2}$ , and $𝑎 + 𝑑 = tr 𝐴$ .

(b) Here $tr 𝐴 = 0$ and $det 𝐴 =$ $0 \cdot$ $0 -$ $(- 1) \cdot$ $1 =$ $1$ , so $det (𝐴 + 𝑡 𝐼_{2}) =$ $𝑡^{2} +$ $1$ . By Exercise 5.1.25, $𝐴 + 𝑡 𝐼_{2}$ fails to be invertible exactly when $𝑡^{2} + 1 = 0$ . Over $ℂ$ this means $𝑡 = 𝑖$ or $𝑡 = - 𝑖$ (Exercise 1.1.44). Over $ℝ$ there is no such $𝑡$ , since $𝑡^{2} + 1 \geq 1$ .

So $𝐴 𝐱 = - 𝑡 𝐱$ has a nonzero solution $𝐱$ for $𝑡 = \pm 𝑖$ over $ℂ$ but for no real $𝑡$ : the rotation by $𝜋 / 2$ moves every nonzero real vector off its own line. This is the first appearance of eigenvalues, which Section 7.3 finds through exactly this kind of determinant.

Exercise 5.1.39.standard Let $𝑝$ be a prime.

(a)Count the matrices $𝐴 \in 𝔽_{𝑝}^{2 \times 2}$ with $det 𝐴 \neq 0$ .
(b)For $𝑝 = 3$ , show that exactly half of them have $det 𝐴 = 1$ .
(c)How many $𝐴 \in 𝔽_{2}^{2 \times 2}$ have $det 𝐴 = 1$ ?

Solution.

Uses. Exercise 5.1.25, Definition 1.5.1 (Linear independence), Proposition 5.1.10 (Alternating implies antisymmetric), Proposition 1.2.35 (Zero products)

Hint 1. By Exercise 5.1.25, count the pairs of columns that are linearly independent. Choose the first column, then the second.

Hint 2. Two vectors are independent exactly when the first is nonzero and the second is not a multiple of the first. How many multiples does a nonzero vector of

𝔽_{𝑝}^{2}

have?

Hint 3. For (b), interchanging the columns is a bijection on

𝔽_{3}^{2 \times 2}

. What does it do to the determinant?

Sketch. (a)

(𝑝^{2} - 1) (𝑝^{2} - 𝑝)

: a nonzero first column, then a second column outside its

𝑝

multiples. (b) For

𝑝 = 3

that is

48

; interchanging the columns negates

det

and so swaps

det = 1

with

det = 2 = - 1

. (c)

(4 - 1) (4 - 2) = 6

, all with

det = 1

(a) By Exercise 5.1.25, $det 𝐴 \neq 0$ exactly when the columns $𝐚_{1}, 𝐚_{2}$ are linearly independent, which happens exactly when $𝐚_{1} \neq 𝟎$ and $𝐚_{2}$ is not a multiple of $𝐚_{1}$ . (If $𝐚_{1} = 𝟎$ , then $1 \cdot 𝐚_{1} = 𝟎$ ; if $𝐚_{2} = 𝑐 𝐚_{1}$ , then $𝑐 𝐚_{1} - 𝐚_{2} = 𝟎$ . Conversely, if $𝑥 𝐚_{1} + 𝑦 𝐚_{2} = 𝟎$ with $𝑦 \neq 0$ , then $𝐚_{2} = - 𝑦^{- 1} 𝑥 𝐚_{1}$ , and if $𝑦 = 0$ then $𝑥 𝐚_{1} = 𝟎$ with $𝑥 \neq 0$ , so $𝐚_{1} = 𝟎$ .) There are $𝑝^{2} - 1$ nonzero vectors in $𝔽_{𝑝}^{2}$ . For $𝐚_{1} \neq 𝟎$ , the multiples $𝑐 𝐚_{1}$ with $𝑐 \in 𝔽_{𝑝}$ are pairwise distinct, because $𝑐 𝐚_{1} = 𝑐^{'} 𝐚_{1}$ gives $(𝑐 - 𝑐^{'}) 𝐚_{1} = 𝟎$ and hence $𝑐 = 𝑐^{'}$ (Proposition 1.2.35). So there are $𝑝^{2} - 𝑝$ choices for $𝐚_{2}$ , and $(𝑝^{2} - 1) (𝑝^{2} - 𝑝)$ matrices in all.

(b) For $𝑝 = 3$ the count is $8 \cdot 6 = 48$ . The nonzero elements of $𝔽_{3}$ are $1$ and $2 = - 1$ . Interchanging the two columns is a bijection from $𝔽_{3}^{2 \times 2}$ to itself (it is its own inverse), and it negates the determinant (Proposition 5.1.10). So it maps the matrices with $det 𝐴 = 1$ bijectively onto those with $det 𝐴 = - 1 = 2$ . These two sets have the same size and together contain all $48$ matrices, so each has $24$ elements.

(c) For $𝑝 = 2$ the count is $(4 - 1) (4 - 2) = 6$ , and the only nonzero element of $𝔽_{2}$ is $1$ , so all six have $det 𝐴 = 1$ .

Exercise 5.1.40.hard Let $𝑛 \geq 2$ , and let $𝐷 : 𝐹^{𝑛 \times 𝑛} \to 𝐹$ be multilinear and alternating. Write $𝐚_{𝑛 + 1} = 𝐚_{1}$ .

(a)Show that when $𝑛 = 3$ ,
$𝐷 (𝐚_{1} + 𝐚_{2}, 𝐚_{2} + 𝐚_{3}, 𝐚_{3} + 𝐚_{1}) = 2 𝐷 (𝐚_{1}, 𝐚_{2}, 𝐚_{3}) .$
(b)Show that in general
$𝐷 (𝐚_{1} + 𝐚_{2}, 𝐚_{2} + 𝐚_{3}, \dots, 𝐚_{𝑛} + 𝐚_{𝑛 + 1}) = (1 + {(- 1)}^{𝑛 - 1}) 𝐷 (𝐚_{1}, \dots, 𝐚_{𝑛}) .$
(c)For even $𝑛$ , explain the answer to (b) directly by finding a linear dependence among the columns $𝐚_{𝑘} + 𝐚_{𝑘 + 1}$ .

Solution.

Uses. Proposition 5.1.10 (Alternating implies antisymmetric), Proposition 5.1.17 (Dependent columns)

Techniques. Expand by multilinearity (Technique 5.1.7)

Hint 1. Expand every column. A term corresponds to choosing

𝐚_{𝑘}

𝐚_{𝑘 + 1}

in each column

𝑘

, so there are

2^{𝑛}

terms.

Hint 2. Show that if the choices are not all “first” or all “second”, then some vector is chosen twice. Look for a column

𝑘

where the second option is chosen, followed by a column

𝑘 + 1

where the first is.

Hint 3. The all-second term is

𝐷 (𝐚_{2}, \dots, 𝐚_{𝑛}, 𝐚_{1})

. Move

𝐚_{1}

back to the front by adjacent interchanges.

Sketch. Only the two constant choices survive: a switch from “second” in column

𝑘

to “first” in column

𝑘 + 1

puts

𝐚_{𝑘 + 1}

in both columns. So the value is

𝐷 (𝐚_{1}, \dots, 𝐚_{𝑛}) +

𝐷 (𝐚_{2}, \dots, 𝐚_{𝑛}, 𝐚_{1})

, and

𝑛 - 1

adjacent interchanges give the second term as

{(- 1)}^{𝑛 - 1} 𝐷 (𝐚_{1}, \dots, 𝐚_{𝑛})

. For even

𝑛

, the alternating sum of the columns telescopes to

𝟎

We prove (b); (a) is the case $𝑛 = 3$ , where $1 + {(- 1)}^{2} = 2$ . Indices are read modulo $𝑛$ , so column $𝑛$ is $𝐚_{𝑛} + 𝐚_{1}$ .

(b) Expanding each column $𝑘$ into its two terms writes the left side as a sum of $2^{𝑛}$ terms, one for each $𝜀 =$ $(𝜀_{1}, \dots, 𝜀_{𝑛}) \in$ ${0, 1}^{𝑛}$ :

𝑇_{𝜀} = 𝐷 (𝐚_{1 + 𝜀_{1}}, 𝐚_{2 + 𝜀_{2}}, \dots, 𝐚_{𝑛 + 𝜀_{𝑛}}) .

Suppose $𝜀$ is not constant. Choose $𝑘$ with $𝜀_{𝑘} = 1$ , and walk forward cyclically from $𝑘$ . Since some entry of $𝜀$ is $0$ , there is a first index $𝑙$ after $𝑘$ (cyclically) with $𝜀_{𝑙} = 0$ , and then $𝜀_{𝑙 - 1} = 1$ . So column $𝑙 - 1$ of $𝑇_{𝜀}$ holds $𝐚_{𝑙}$ , and column $𝑙$ also holds $𝐚_{𝑙}$ . These are different columns because $𝑛 \geq 2$ , so $𝑇_{𝜀} = 0$ since $𝐷$ is alternating. Only the two constant choices remain:

𝐷 (𝐚_{1} + 𝐚_{2}, \dots, 𝐚_{𝑛} + 𝐚_{1}) = 𝐷 (𝐚_{1}, 𝐚_{2}, \dots, 𝐚_{𝑛}) + 𝐷 (𝐚_{2}, \dots, 𝐚_{𝑛}, 𝐚_{1}) .

In the second term, interchange columns $𝑛 - 1$ and $𝑛$ , then columns $𝑛 - 2$ and $𝑛 - 1$ , and so on, ending with columns $1$ and $2$ . These $𝑛 - 1$ interchanges move $𝐚_{1}$ from the last position to the first and keep the others in order, and each negates $𝐷$ (Proposition 5.1.10). So the second term is ${(- 1)}^{𝑛 - 1} 𝐷 (𝐚_{1}, \dots, 𝐚_{𝑛})$ , which proves (b).

(c) Let $𝑛$ be even and $𝐜_{𝑘} = 𝐚_{𝑘} + 𝐚_{𝑘 + 1}$ . Then the sum telescopes:

𝐜_{1} - 𝐜_{2} + 𝐜_{3} - \dots - 𝐜_{𝑛} = 𝐚_{1} + (𝐚_{2} - 𝐚_{2}) + \dots + (𝐚_{𝑛} - 𝐚_{𝑛}) - 𝐚_{𝑛 + 1} = 𝐚_{1} - 𝐚_{1} = 𝟎 .

The coefficients are $\pm 1 \neq 0$ , so the columns are dependent, and the left side of (b) is $0$ by Proposition 5.1.17, matching $1 + {(- 1)}^{𝑛 - 1} = 0$ .

Exercise 5.1.41.hard Let $𝐷 : 𝐹^{𝑛 \times 𝑛} \to 𝐹$ satisfy the following two conditions.

(i)Scaling a column scales $𝐷$ : $𝐷 (\dots, 𝑐 𝐱, \dots) =$ $𝑐 𝐷 (\dots, 𝐱, \dots)$ for every column position, every $𝑐 \in 𝐹$ , and all fixed other columns.
(ii)Column replacement preserves $𝐷$ : if $𝐴^{'}$ is obtained from $𝐴$ by adding a multiple of one column to a different column, then $𝐷 (𝐴^{'}) = 𝐷 (𝐴)$ .

Show that $𝐷$ is multilinear and alternating. Since conversely every multilinear alternating function satisfies (i) and (ii) (by Lemma 5.1.8), multilinearity and alternation may be replaced by (i) and (ii) in the definition of a determinant function. These are the two properties of volume that are visible in the shear picture.

Solution.

Uses. Theorem 4.1.16 (Extending to a basis), Definition 4.1.5 (Dimension), Proposition 1.6.19 (Coordinates respect linear combinations), Definition 1.5.1 (Linear independence)

Hint 1. Alternation: if two columns are equal, use (ii) to turn one of them into

𝟎

, then use (i) with

𝑐 = 0

Hint 2. Additivity in column

𝑘

: fix the other

𝑛 - 1

columns. If they are dependent, show

𝐷

vanishes for every choice of column

𝑘

Hint 3. If they are independent, extend them to a basis of

𝐹^{𝑛}

by one vector

𝐰

. Write

𝐱 = 𝑠 𝐰 + (\dots)

and use (ii) to strip off the other part.

Sketch. Subtracting one of two equal columns from the other makes a zero column, and (i) with

𝑐 = 0

gives

𝐷 = 0

. Fix all columns but the

𝑘

-th. If they are dependent, (ii) produces a zero column among them, so

𝐷 (\dots, 𝐱, \dots) = 0

for all

𝐱

. Otherwise extend them by one vector

𝐰

to a basis; writing

𝐱 =

𝑠 (𝐱) 𝐰 +

\sum_{𝑗 \neq 𝑘} 𝑐_{𝑗} 𝐚_{𝑗}

, (ii) and (i) give

𝐷 (\dots, 𝐱, \dots) =

𝑠 (𝐱) 𝐷 (\dots, 𝐰, \dots)

, where

𝑠

is a coordinate function and hence linear.

A zero column. If column $𝑘$ of $𝐴$ is $𝟎$ , then by (i) with $𝑐 = 0$ , $𝐷 (𝐴) =$ $𝐷 (\dots, 0 \cdot 𝟎, \dots) =$ $0 \cdot$ $𝐷 (𝐴) =$ $0$ .

Alternating. Suppose $𝐚_{𝑖} = 𝐚_{𝑗}$ with $𝑖 \neq 𝑗$ . Adding $- 1$ times column $𝑗$ to column $𝑖$ does not change $𝐷$ , by (ii), and produces a matrix whose column $𝑖$ is $𝟎$ . So $𝐷 (𝐴) = 0$ .

Multilinear. Fix $𝑘$ and the columns $𝐚_{𝑗}$ for $𝑗 \neq 𝑘$ , and let $𝑔 (𝐱)$ be the value of $𝐷$ with $𝐱$ in column $𝑘$ . We show that $𝑔$ is linear.

Suppose first that the list ${(𝐚_{𝑗})}_{𝑗 \neq 𝑘}$ is linearly dependent (so $𝑛 \geq 2$ ). Then some $𝐚_{𝑚}$ with $𝑚 \neq 𝑘$ is a linear combination $\sum 𝑏_{𝑗} 𝐚_{𝑗}$ of the others, the sum running over $𝑗 \notin {𝑘, 𝑚}$ . For any $𝐱$ , adding $- 𝑏_{𝑗}$ times column $𝑗$ to column $𝑚$ for each such $𝑗$ does not change $𝐷$ , by (ii), and turns column $𝑚$ into $𝟎$ . So $𝑔 (𝐱) = 0$ for every $𝐱$ , and $𝑔$ is linear.

Suppose instead that ${(𝐚_{𝑗})}_{𝑗 \neq 𝑘}$ is linearly independent. It has $𝑛 - 1$ vectors, so by Theorem 4.1.16 it extends to a basis of $𝐹^{𝑛}$ , and since every basis of $𝐹^{𝑛}$ has $𝑛$ vectors (Definition 4.1.5), exactly one vector $𝐰$ is added. Let $𝑠 (𝐱)$ be the coefficient of $𝐰$ when $𝐱$ is written in this basis, so $𝐱 =$ $𝑠 (𝐱) 𝐰 +$ $\sum_{𝑗 \neq 𝑘} 𝑐_{𝑗} 𝐚_{𝑗}$ for some scalars $𝑐_{𝑗}$ . The function $𝑠$ is a coordinate of the coordinate map, so it is linear (Proposition 1.6.19). Adding $- 𝑐_{𝑗}$ times column $𝑗$ to column $𝑘$ for each $𝑗 \neq 𝑘$ does not change $𝐷$ , by (ii), so by (i)

𝑔 (𝐱) = 𝑔 (𝑠 (𝐱) 𝐰) = 𝑠 (𝐱) 𝑔 (𝐰) .

So $𝑔$ is the linear function $𝑠$ times the constant $𝑔 (𝐰)$ , hence linear. In both cases $𝐷$ is linear in column $𝑘$ , and $𝑘$ was arbitrary.

Exercise 5.1.42.challenge Let

𝐷 : ℝ^{2 \times 2} \to ℝ

satisfy

𝐷 (𝐴 𝐵) =

𝐷 (𝐴) 𝐷 (𝐵)

for all

𝐴, 𝐵

, and

𝐷 (\begin{matrix} 𝑡 & 0 \\ 0 & 1 \end{matrix}) = 𝑡

for all

𝑡 \in ℝ

. Show that

𝐷 = det

Solution.

Uses. Theorem 3.5.19 (Invertible matrices are products of elementary matrices), Definition 3.2.2 (Elementary matrices), Exercise 5.1.27, Exercise 5.1.25

Hint 1. First show

𝐷 (𝐼_{2}) = 1

, that

𝐷 (𝐺) \neq 0

for invertible

𝐺

, and that

𝐷 (𝐺 𝑋 𝐺^{- 1}) =

𝐷 (𝑋)

Hint 2. For the shear

𝑆_{𝑡} = (\begin{matrix} 1 & 𝑡 \\ 0 & 1 \end{matrix})

, check that

(\begin{matrix} 𝑡 & 0 \\ 0 & 1 \end{matrix}) 𝑆_{1} {(\begin{matrix} 𝑡 & 0 \\ 0 & 1 \end{matrix})}^{- 1} =

𝑆_{𝑡}

and that

𝑆_{1}^{2} = 𝑆_{2}

. Deduce

𝐷 (𝑆_{1}) = 1

Hint 3. Conjugating by the swap

𝑃 = (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})

handles lower shears and

(\begin{matrix} 1 & 0 \\ 0 & 𝑡 \end{matrix})

. For

𝑃

itself, check that

𝑆_{1} (\begin{matrix} 1 & 0 \\ - 1 & 1 \end{matrix}) 𝑆_{1} (\begin{matrix} - 1 & 0 \\ 0 & 1 \end{matrix}) =

𝑃

Hint 4. Now

𝐷 (𝐸) = det 𝐸

for every elementary matrix. Use Theorem 3.5.19 for invertible

𝐴

. If

𝐴

is not invertible, write a zero column with a shear and use

𝐷 (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}) = 0

Sketch.

𝐷

is invariant under conjugation by invertible matrices. Conjugating

𝑆_{1}

(\begin{matrix} 𝑡 & 0 \\ 0 & 1 \end{matrix})

gives

𝑆_{𝑡}

, so

𝐷 (𝑆_{1}) =

𝐷 (𝑆_{2}) =

{𝐷 (𝑆_{1})}^{2}

, and

𝐷 (𝑆_{1}) = 1

. Conjugating by

𝑃

gives the lower shears and the second scaling, and

𝑃

is three shears times

(\begin{matrix} - 1 & 0 \\ 0 & 1 \end{matrix})

, so

𝐷 (𝑃) = - 1

. So

𝐷 = det

on elementary matrices, hence on invertible matrices. A non-invertible

𝐴

satisfies

𝐴 𝑆 = 𝐴 𝑆 (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix})

𝐴 = 𝐴 (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix})

for a shear

𝑆

, so

𝐷 (𝐴) = 0

Write $diag (𝑠, 𝑡) =$ $(\begin{matrix} 𝑠 & 0 \\ 0 & 𝑡 \end{matrix})$ , $𝑆_{𝑡} = (\begin{matrix} 1 & 𝑡 \\ 0 & 1 \end{matrix})$ , $𝐿_{𝑡} = (\begin{matrix} 1 & 0 \\ 𝑡 & 1 \end{matrix})$ , and $𝑃 = (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})$ .

Step 0. $𝐷 (𝐼_{2}) =$ $𝐷 (diag (1, 1)) =$ $1$ . If $𝐺$ is invertible, then $𝐷 (𝐺) 𝐷 (𝐺^{- 1}) =$ $𝐷 (𝐼_{2}) =$ $1$ , so $𝐷 (𝐺) \neq 0$ and $𝐷 (𝐺^{- 1}) = {𝐷 (𝐺)}^{- 1}$ . Hence $𝐷 (𝐺 𝑋 𝐺^{- 1}) =$ $𝐷 (𝐺) 𝐷 (𝑋) {𝐷 (𝐺)}^{- 1} =$ $𝐷 (𝑋)$ for every $𝑋$ .

Step 1: shears. For $𝑡 \neq 0$ , a direct multiplication gives $diag (𝑡, 1) 𝑆_{1} diag (𝑡^{- 1}, 1) =$ $𝑆_{𝑡}$ , so $𝐷 (𝑆_{𝑡}) = 𝐷 (𝑆_{1})$ by Step 0. Also $𝑆_{1} 𝑆_{1} = 𝑆_{2}$ , so ${𝐷 (𝑆_{1})}^{2} =$ $𝐷 (𝑆_{2}) =$ $𝐷 (𝑆_{1})$ . Since $𝑆_{1}$ is invertible, $𝐷 (𝑆_{1}) \neq 0$ , and therefore $𝐷 (𝑆_{1}) = 1$ . So $𝐷 (𝑆_{𝑡}) = 1$ for all $𝑡$ (for $𝑡 = 0$ , $𝑆_{0} = 𝐼_{2}$ ). Next, $𝑃 = 𝑃^{- 1}$ and $𝑃 𝑆_{𝑡} 𝑃 = 𝐿_{𝑡}$ , so $𝐷 (𝐿_{𝑡}) = 1$ as well.

Step 2: scalings and the swap. $𝑃 diag (𝑡, 1) 𝑃 =$ $diag (1, 𝑡)$ , so $𝐷 (diag (1, 𝑡)) = 𝑡$ . Multiplying out,

𝑆_{1} 𝐿_{- 1} 𝑆_{1} = (\begin{matrix} 0 & 1 \\ - 1 & 1 \end{matrix}) 𝑆_{1} = (\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix}), (\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix}) diag (- 1, 1) = 𝑃,

so $𝐷 (𝑃) =$ $1 \cdot$ $1 \cdot$ $1 \cdot$ $(- 1) =$ $- 1$ .

Step 3: invertible matrices. The elementary matrices in $ℝ^{2 \times 2}$ are $𝑃$ , the scalings $diag (𝑡, 1)$ and $diag (1, 𝑡)$ with $𝑡 \neq 0$ , and the shears $𝑆_{𝑡}$ and $𝐿_{𝑡}$ . By Steps 1 and 2, $𝐷 (𝐸) = det 𝐸$ for each of them ( $det 𝑃 = - 1$ , $det diag (𝑡, 1) =$ $det diag (1, 𝑡) =$ $𝑡$ , and $det 𝑆_{𝑡} = det 𝐿_{𝑡} = 1$ ). If $𝐴$ is invertible, then $𝐴 = 𝐸_{1} \dots 𝐸_{𝑘}$ for some elementary matrices $𝐸_{𝑖}$ (Theorem 3.5.19), and by multiplicativity of $𝐷$ and of $det$ (Exercise 5.1.27),

𝐷 (𝐴) = 𝐷 (𝐸_{1}) \dots 𝐷 (𝐸_{𝑘}) = det (𝐸_{1}) \dots det (𝐸_{𝑘}) = det 𝐴 .

Step 4: non-invertible matrices. Let $𝐴$ be non-invertible, so $det 𝐴 = 0$ and its columns are dependent (Exercise 5.1.25). Note $𝐷 (diag (0, 1)) = 0$ , and $𝐷 (diag (1, 0)) =$ $𝐷 (𝑃 diag (0, 1) 𝑃) =$ $0$ by Step 0. If $𝐚_{1} = 𝟎$ , then $𝐴 = 𝐴 diag (0, 1)$ , so $𝐷 (𝐴) = 𝐷 (𝐴) \cdot 0 = 0$ . If $𝐚_{1} \neq 𝟎$ , a dependence $𝑥 𝐚_{1} + 𝑦 𝐚_{2} = 𝟎$ must have $𝑦 \neq 0$ , so $𝐚_{2} = 𝑐 𝐚_{1}$ with $𝑐 = - 𝑦^{- 1} 𝑥$ . Then $𝐴 𝑆_{- 𝑐}$ has columns $𝐚_{1}$ and $𝟎$ , so $𝐴 𝑆_{- 𝑐} =$ $𝐴 𝑆_{- 𝑐} diag (1, 0)$ , and

𝐷 (𝐴) = 𝐷 (𝐴) 𝐷 (𝑆_{- 𝑐}) = 𝐷 (𝐴 𝑆_{- 𝑐}) = 𝐷 (𝐴 𝑆_{- 𝑐}) 𝐷 (diag (1, 0)) = 0 .

In both cases $𝐷 (𝐴) = 0 = det 𝐴$ . Hence $𝐷 = det$ .

More practice

The practice bank has 30 exercises on this section’s topics.

5.1.1Signed area

5.1.2The determinant axioms

5.1.3Swapping columns

5.1.4The 2×2 determinant

5.1.5Dependent columns

Exercises

More practice

5.1.4The $2 \times 2$ determinant