Problems for Chapter 5

(a) Statement (Theorem 5.3.4). Let $𝑛\ge 1$. The function

is a determinant function on ${𝐹}^{𝑛\times 𝑛}$, and it is the only one. Moreover, every multilinear alternating $𝐷:{𝐹}^{𝑛\times 𝑛}\to 𝐹$ satisfies $𝐷=𝐷({𝐼}_{𝑛})𝐿$.

(b) Let $𝐴$ have equal columns $𝑘$ and $𝑙$, where $𝑘\ne 𝑙$, and let $𝜏=(𝑘𝑙)$. Write ${𝑡}_{𝜎}(𝐴)=$$sgn(𝜎){\prod }_{𝑗}{𝑎}_{𝜎(𝑗)𝑗}$. The permutation $𝜎𝜏$ agrees with $𝜎$ except that $(𝜎𝜏)(𝑘)=𝜎(𝑙)$ and $(𝜎𝜏)(𝑙)=𝜎(𝑘)$. So

and since columns $𝑘$ and $𝑙$ are equal, ${𝑎}_{𝜎(𝑙)𝑘}=$${𝑎}_{𝜎(𝑙)𝑙}$ and ${𝑎}_{𝜎(𝑘)𝑙}=$${𝑎}_{𝜎(𝑘)𝑘}$. So the products of entries in ${𝑡}_{𝜎𝜏}(𝐴)$ and ${𝑡}_{𝜎}(𝐴)$ agree, while $sgn(𝜎𝜏)=$$-sgn(𝜎)$ by Proposition 5.2.25. Hence ${𝑡}_{𝜎𝜏}(𝐴)=$$-{𝑡}_{𝜎}(𝐴)$.

The map $𝜎\mapsto 𝜎𝜏$ is its own inverse and changes the sign, so it maps the even permutations bijectively onto the odd ones. Therefore

No step divides by $1+1$: over ${𝔽}_2$ the two terms of a pair are equal and cancel because $𝑥+𝑥=0$.

(c) Write $𝑔(𝐴)=$${𝑎}_{11}{𝑎}_{12}\cdots {𝑎}_{1𝑛}$, so $𝐷=det+𝑔$.

Normalized. $det{𝐼}_{𝑛}=1$, and $𝑔({𝐼}_{𝑛})$ contains the factor ${𝑎}_{12}=0$ (as $𝑛\ge 2$). So $𝐷({𝐼}_{𝑛})=1$.

Multilinear. $𝑔$ contains exactly one entry of each column, namely ${𝑎}_{1𝑘}$ from column $𝑘$. With the other columns fixed, $𝑔$ is a constant times ${𝑎}_{1𝑘}$, a linear function of column $𝑘$. So $𝑔$ is multilinear, and so is $𝐷=det+𝑔$.

Antisymmetric. Interchanging two columns negates $det$ (Proposition 5.1.10) and permutes the factors of $𝑔$, so it leaves $𝑔$ unchanged. In ${𝔽}_2$, $-1=1$, so both $det$ and $𝑔$ are unchanged, and $𝐷({𝐴}^{\prime })=$$𝐷(𝐴)=$$-𝐷(𝐴)$ for the interchanged matrix ${𝐴}^{\prime }$.

Not alternating. Let $𝐴$ have first row $(1,\dots ,1)$ and all other rows $0$. All columns of $𝐴$ are equal to ${𝐞}_1$, so $det𝐴=0$, but $𝑔(𝐴)=1$, and $𝐷(𝐴)=1\ne 0$.

If $1+1\ne 0$, an antisymmetric function is alternating (Proposition 5.1.11). So a normalized, multilinear, antisymmetric $𝐷$ would be a determinant function, hence equal to $det$ by uniqueness, and $det$ is alternating. So over such a field the three conditions determine $det$, while over ${𝔽}_2$ they allow at least two functions, for every $𝑛\ge 2$. This is why the definition asks for alternation (compare Example 5.1.12, the case $𝑛=2$).

(a) Theorem 5.4.7: $𝐴\in {𝐹}^{𝑛\times 𝑛}$ is invertible if and only if $det𝐴\ne 0$. Theorem 5.4.10: $det(𝐴𝐵)=$$det𝐴det𝐵$ for all $𝐴,𝐵\in {𝐹}^{𝑛\times 𝑛}$.

(b) Fix $𝐴$, and let $𝐷(𝐵)=det(𝐴𝐵)$. Column $𝑗$ of $𝐴𝐵$ is $𝐴{𝐛}_{𝑗}$, so $𝐷(𝐵)=$$det(𝐴{𝐛}_1,\dots ,𝐴{𝐛}_{𝑛})$. In column $𝑗$, $𝐷$ is the linear map $𝐱\mapsto 𝐴𝐱$ followed by $det$, which is linear in column $𝑗$; so $𝐷$ is multilinear. If two columns of $𝐵$ are equal, the corresponding columns of $𝐴𝐵$ are equal, and $𝐷(𝐵)=0$; so $𝐷$ is alternating. By Theorem 5.3.4, $𝐷(𝐵)=$$𝐷({𝐼}_{𝑛})det𝐵=$$det(𝐴)det𝐵$.

If $𝐴$ is invertible, then $1=$$det{𝐼}_{𝑛}=$$det(𝐴{𝐴}^{-1})=$$det𝐴det{𝐴}^{-1}$, so $det𝐴\ne 0$.

(c) Suppose $𝐴\in {\mathbb{Q}}^{𝑛\times 𝑛}$ and ${𝐴}^2=2{𝐼}_{𝑛}$. By (b) and Exercise 5.1.5, ${(det𝐴)}^2=$$det(2{𝐼}_{𝑛})=$$2^{𝑛}$, and $det𝐴\in \mathbb{Q}$. If $𝑛=2𝑘+1$ is odd, then ${(det𝐴/2^{𝑘})}^2=$$2^{2𝑘+1}/2^{2𝑘}=$$2$, so $\sqrt{2}$ would be rational, which it is not. So $𝑛$ must be even. Conversely, $𝑅=\left(\begin{array}{cc}0& 2\\ 1& 0\end{array}\right)$ satisfies ${𝑅}^2=2{𝐼}_2$, and for even $𝑛$ the block-diagonal matrix with $𝑛/2$ blocks $𝑅$ satisfies ${𝐴}^2=2{𝐼}_{𝑛}$, since block-diagonal matrices multiply block by block (Theorem 2.4.34). So the answer over $\mathbb{Q}$ is: exactly the even $𝑛$.

Over $ℝ$, $𝐴=\sqrt{2}{𝐼}_{𝑛}$ works for every $𝑛$. The obstruction over $\mathbb{Q}$ is arithmetic, the irrationality of $\sqrt{2}$, and the determinant is what detects it.

(a) (Theorem 5.5.14) If $𝐵\in {𝐹}^{𝑛\times 𝑛}$ has $det𝐵\ne 0$ and $𝐲\in {𝐹}^{𝑛}$, then $𝐵𝐱=𝐲$ has exactly one solution, and ${𝑥}_{𝑖}=$$det{𝐵}_{𝑖}(𝐲)/det𝐵$ for each $𝑖$.

(b) Let $𝐱$ be a solution. Then $𝐲=$$𝐵𝐱=$${\sum }_{𝑘}{𝑥}_{𝑘}{𝐛}_{𝑘}$ (Proposition 2.4.8). By linearity in column $𝑖$,

For $𝑘\ne 𝑖$, the matrix ${𝐵}_{𝑖}({𝐛}_{𝑘})$ has ${𝐛}_{𝑘}$ in columns $𝑖$ and $𝑘$, so its determinant is $0$. For $𝑘=𝑖$, ${𝐵}_{𝑖}({𝐛}_{𝑖})=𝐵$. Hence $det{𝐵}_{𝑖}(𝐲)=$${𝑥}_{𝑖}det𝐵$.

If $det𝐵\ne 0$, then $𝐵$ is invertible (Theorem 5.4.7), so $𝐱={𝐵}^{-1}𝐲$ is the unique solution, and dividing by $det𝐵$ gives ${𝑥}_{𝑖}=$$det{𝐵}_{𝑖}(𝐲)/det𝐵$.

(c) If $𝐲$ is integral, then ${𝐵}_{𝑖}(𝐲)$ has integer entries, so $det{𝐵}_{𝑖}(𝐲)$ is an integer (Proposition 5.3.20), and by (b), $𝑑{𝑥}_{𝑖}=$$det{𝐵}_{𝑖}(𝐲)\in$$\mathbb{Z}$.

Suppose that every integral $𝐲$ gives an integral solution. Then each column ${𝐵}^{-1}{𝐞}_{𝑗}$ of ${𝐵}^{-1}$, the solution for $𝐲={𝐞}_{𝑗}$, is integral, so ${𝐵}^{-1}$ has integer entries, and $𝑑=\pm 1$ by Corollary 5.5.12. So if $𝑑\ne \pm 1$, some ${𝐞}_{𝑗}$ has a non-integral solution.

For $𝐵=\left(\begin{array}{cc}2& 1\\ 1& 3\end{array}\right)$, $𝑑=5$. With $𝐲={𝐞}_1$,

Check: $2\cdot 3/5-1/5=1$ and $3/5+3\cdot (-1/5)=0$. The denominators are exactly $𝑑$.

(a) False. Let $𝐴={𝐼}_{𝑛}$ and $𝐵=$$diag(-1,-1,1,\dots ,1)$. Both have determinant $1$ (Proposition 5.3.15), but $𝐴+$$𝐵=$$diag(0,0,2,\dots ,2)$ has determinant $0$.

(b) True. Let $𝑑=det𝐴$. By Theorem 5.4.10, ${𝑑}^3=$$det({𝐴}^3)=$$det𝐴=$$𝑑$, so $𝑑(𝑑-1)(𝑑+1)=$${𝑑}^3-$$𝑑=$$0$. A field has no zero divisors (Proposition 1.1.17), so $𝑑$ is $0$, $1$, or $-1$. (Over ${𝔽}_2$ the set $\{-1,0,1\}$ is $\{0,1\}$.)

(c) False in general. Let $𝐴=\left(\begin{array}{cc}1& 0\end{array}\right)\in {ℝ}^{1\times 2}$. Then ${𝐴}^{\text{𝖳}}𝐴=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)$ has determinant $0$, while $𝐴{𝐴}^{\text{𝖳}}=(1)$ has determinant $1$. The statement is true when $𝑚=𝑛$: then both sides equal $det{𝐴}^{\text{𝖳}}det𝐴=$${(det𝐴)}^2$ by Theorem 5.4.10 and Theorem 5.4.14.

(d) True. The $𝑖$-th entry of $𝐴(1,\dots ,1)$ is the sum of row $𝑖$, so $𝐴(1,\dots ,1)=\mathrm{𝟎}$ with $(1,\dots ,1)\ne \mathrm{𝟎}$, and $det𝐴=0$ by Corollary 5.4.8.

(e) False for every $𝑛\ge 2$ (for $𝑛=1$ it is true). $\left(\begin{array}{cc}1& 0\\ 1& 0\end{array}\right)$ has row sums $1$ and determinant $0$, and adding a diagonal block ${𝐼}_{𝑛-2}$ gives an example for each $𝑛\ge 2$. The hypothesis says $𝐴(1,\dots ,1)=(1,\dots ,1)$, a statement about one vector, which says nothing about $det𝐴$.

(a) By Definition 5.1.14 we count the triples $(𝑎,𝑏,𝑑)\in {𝔽}_{𝑝}^3$ with $𝑎𝑑-{𝑏}^2=0$. If $𝑏=0$, we need $𝑎𝑑=0$, that is, $𝑎=0$ or $𝑑=0$ (Proposition 1.1.17): there are $𝑝$ triples with $𝑎=0$ and $𝑝$ with $𝑑=0$, and one triple has both, so $2𝑝-1$ in all. If $𝑏\ne 0$ ($𝑝-1$ choices), then $𝑎𝑑={𝑏}^2\ne 0$, so $𝑎\ne 0$ ($𝑝-1$ choices), and then $𝑑={𝑏}^2{𝑎}^{-1}$ is determined. In all there are

such matrices.

(b) Write $𝐴=\left(\begin{array}{cc}𝑎& 𝑏\\ 𝑐& 𝑑\end{array}\right)$. Then $tr𝐴=0$ means $𝑑=-𝑎$, and then $det𝐴=-{𝑎}^2-𝑏𝑐$, so the conditions say $𝑏(-𝑐)={𝑎}^2$, with $𝑎,𝑏,𝑐$ free otherwise. The map $(𝑎,𝑏,𝑐)\mapsto$$(𝑏,-𝑐,𝑎)$ is a bijection of ${𝔽}_{𝑝}^3$ that turns these solutions into the solutions $(𝑥,𝑦,𝑧)$ of $𝑥𝑦={𝑧}^2$, and so does $(𝑎,𝑏,𝑑)\mapsto$$(𝑎,𝑑,𝑏)$ for the equation in (a). Both counts are the number of solutions of $𝑥𝑦={𝑧}^2$ in ${𝔽}_{𝑝}^3$, which is ${𝑝}^2$.

(c) With $𝐴=\left(\begin{array}{cc}𝑎& 𝑏\\ 𝑐& 𝑑\end{array}\right)$,

so ${𝐴}^2-$$(tr𝐴)𝐴=$$\left(\begin{array}{cc}𝑏𝑐-𝑎𝑑& 0\\ 0& 𝑏𝑐-𝑎𝑑\end{array}\right)=$$-(det𝐴){𝐼}_2$, as claimed.

If $tr𝐴=det𝐴=0$, the identity gives ${𝐴}^2=0$. Conversely, suppose ${𝐴}^2=0$. Then ${(det𝐴)}^2=$$det({𝐴}^2)=$$0$ by Theorem 5.4.10, so $det𝐴=0$, and the identity becomes $(tr𝐴)𝐴={𝐴}^2=0$. Either $tr𝐴=0$, or $𝐴=0$, and the zero matrix also has trace $0$. So ${𝐴}^2=0$ exactly when $tr𝐴=det𝐴=0$, and by (b) there are ${𝑝}^2$ such matrices in ${𝔽}_{𝑝}^{2\times 2}$, one of which is $0$.

(a) By multilinearity, with $det(𝐜,𝐜)=0$,

Let $𝐬={\sum }_{𝑘=1}^{𝑚}{𝐩}_{𝑘}$. Since ${𝐩}_{𝑚+1}={𝐩}_1$, also ${\sum }_{𝑘=1}^{𝑚}{𝐩}_{𝑘+1}=𝐬$. Summing over $𝑘$ and using linearity in each column, the extra terms add up to

by antisymmetry (Proposition 5.1.10). So $𝑆$ is unchanged.

(b) By (a) with $𝐜=-𝐩$, and writing ${𝐪}^{\prime }=𝐪-𝐩$ and ${𝐫}^{\prime }=𝐫-𝐩$,

since a matrix with a zero column has determinant $0$. The triangle $𝐩𝐪𝐫$ is half of the parallelogram spanned by ${𝐪}^{\prime }$ and ${𝐫}^{\prime }$, translated to $𝐩$. By Example 5.1.15, $det({𝐪}^{\prime },{𝐫}^{\prime })$ is the signed area of that parallelogram, positive exactly when the turn from ${𝐪}^{\prime }$ to ${𝐫}^{\prime }$ is counterclockwise, that is, when $𝐩,𝐪,𝐫$ run counterclockwise.

(c) By (b), $𝑆(𝐜,{𝐩}_{𝑘},{𝐩}_{𝑘+1})=$$\frac{1}{2}det({𝐩}_{𝑘}-𝐜,{𝐩}_{𝑘+1}-𝐜)$. Summing over $𝑘$ gives $𝑆({𝐩}_1-𝐜,\dots ,{𝐩}_{𝑚}-𝐜)$, which is $𝑆({𝐩}_1,\dots ,{𝐩}_{𝑚})$ by (a).

Now let the polygon be convex with counterclockwise vertices, and let $𝐜$ be a point inside it. The segments from $𝐜$ to the vertices cut the polygon into the triangles $𝐜{𝐩}_{𝑘}{𝐩}_{𝑘+1}$, and each of these runs counterclockwise, since the boundary of the polygon does. By (b), each $𝑆(𝐜,{𝐩}_{𝑘},{𝐩}_{𝑘+1})$ is the area of its triangle, and their sum is the area of the polygon.

(d) The terms $det({𝐩}_{𝑘},{𝐩}_{𝑘+1})$ are

so $𝑆=$$\frac{1}{2}(0+12+19+11+0)=$$21$. The pentagon is convex and its vertices run counterclockwise: the edge vectors $(4,0),$$(1,3),$$(-3,2),$$(-3,-2),$$(1,-3)$ point in the directions of about $0°$, $72°$, $146°$, $214°$, and $288°$, so the boundary turns left at every vertex and makes exactly one full turn in all. Hence the area is $21$. (Left turns alone are not enough: the edges of a five-pointed star also turn left at every vertex, but they make two full turns.)

(a) For $𝑖=1,\dots ,𝑛$, subtract column $1$ from column $𝑖+1$. These replacements do not change the determinant (Lemma 5.1.8), and column $𝑖+1$ becomes $(0,{𝐩}_{𝑖}-{𝐩}_0)$. Now row $1$ is $(1,0,\dots ,0)$, so by Lemma 5.5.2 the determinant is ${(-1)}^{1+1}\cdot 1$ times the determinant of the matrix left after deleting row $1$ and column $1$, whose columns are ${𝐩}_1-$${𝐩}_0,$$\dots ,$${𝐩}_{𝑛}-$${𝐩}_0$.

(b) By (a) and Corollary 5.4.8, $det𝑀\ne 0$ exactly when the columns ${𝐩}_{𝑖}-{𝐩}_0$ are independent. For $𝑛=2$ this says that ${𝐩}_1-{𝐩}_0$ and ${𝐩}_2-{𝐩}_0$ do not lie on one line through $\mathrm{𝟎}$, that is, the three points do not lie on one line.

(c) Let $𝛌=({𝜆}_0,\dots ,{𝜆}_{𝑛})$. By Proposition 2.4.8, $𝑀𝛌=$${\sum }_{𝑖}{𝜆}_{𝑖}(1,{𝐩}_{𝑖})$, whose first entry is ${\sum }_{𝑖}{𝜆}_{𝑖}$ and whose other entries form ${\sum }_{𝑖}{𝜆}_{𝑖}{𝐩}_{𝑖}$. So the two conditions say exactly $𝑀𝛌=(1,𝐱)$. Since $det𝑀\ne 0$, Cramer’s rule (Theorem 5.5.14) gives a unique solution, with ${𝜆}_{𝑖}=$$det{𝑀}_{𝑖}/det𝑀$.

(d) ${𝑀}_{𝑖}$ is the matrix $𝑀$ built from the points ${𝐩}_0,\dots ,𝐱,\dots ,{𝐩}_{𝑛}$, with $𝐱$ in place of ${𝐩}_{𝑖}$. By (a) and Problem 5.6 (b), $det𝑀=$$2𝑆({𝐩}_0,{𝐩}_1,{𝐩}_2)$ and

So ${𝜆}_{𝑖}$ is the signed area of the triangle with ${𝐩}_{𝑖}$ replaced by $𝐱$, divided by the signed area of ${𝐩}_0{𝐩}_1{𝐩}_2$. It is positive exactly when the two signed areas have the same sign, that is, the two triangles have the same orientation.

For ${𝐩}_0=(0,0)$, ${𝐩}_1=(4,0)$, ${𝐩}_2=(0,4)$, and $𝐱=(1,2)$:

and $det{𝑀}_0=$$det({𝐩}_1-𝐱,{𝐩}_2-𝐱)=$$det\left(\begin{array}{cc}3& -1\\ -2& 2\end{array}\right)=$$4$. So $𝜆=(1/4,1/4,1/2)$, and indeed $\frac{1}{4}(0,0)+$$\frac{1}{4}(4,0)+$$\frac{1}{2}(0,4)=$$(1,2)$. All three are positive, so $(1,2)$ lies inside the triangle in this sense. These ${𝜆}_{𝑖}$ are the barycentric coordinates of Section 4.7, and the sign test is the standard way to decide whether a point lies in a triangle.

(a) Let ${𝑆}_{𝑖𝑗}=$${𝑈}_{𝑖𝑗}+$${𝑈}_{𝑗𝑖}$ and ${𝐾}_{𝑖𝑗}=$${𝑈}_{𝑖𝑗}-$${𝑈}_{𝑗𝑖}$ for $𝑖<𝑗$. The list ${𝑈}_{11},\dots ,{𝑈}_{𝑛𝑛}$, ${𝑆}_{𝑖𝑗}$, ${𝐾}_{𝑖𝑗}$ has $𝑛+$$2\cdot$$𝑛(𝑛-1)/2=$${𝑛}^2$ entries.

It spans. Each ${𝑈}_{𝑖𝑖}$ is in the list, and for $𝑖<𝑗$, ${𝑈}_{𝑖𝑗}=$${(1+1)}^{-1}({𝑆}_{𝑖𝑗}+{𝐾}_{𝑖𝑗})$ and ${𝑈}_{𝑗𝑖}=$${(1+1)}^{-1}({𝑆}_{𝑖𝑗}-{𝐾}_{𝑖𝑗})$. So the span contains every matrix unit, hence all of ${𝐹}^{𝑛\times 𝑛}$.

It is independent. Suppose ${\sum }_{𝑖}{𝑎}_{𝑖}{𝑈}_{𝑖𝑖}+$${\sum }_{𝑖<𝑗}({𝑏}_{𝑖𝑗}{𝑆}_{𝑖𝑗}+{𝑐}_{𝑖𝑗}{𝐾}_{𝑖𝑗})=$$0$. The $(𝑖,𝑖)$ entry gives ${𝑎}_{𝑖}=0$. For $𝑖<𝑗$, the $(𝑖,𝑗)$ and $(𝑗,𝑖)$ entries give ${𝑏}_{𝑖𝑗}+{𝑐}_{𝑖𝑗}=0$ and ${𝑏}_{𝑖𝑗}-{𝑐}_{𝑖𝑗}=0$, so $(1+1){𝑏}_{𝑖𝑗}=0$, hence ${𝑏}_{𝑖𝑗}=0$ and ${𝑐}_{𝑖𝑗}=0$.

So the list is a basis (Definition 1.6.1). Since ${𝑈}_{𝑖𝑖}^{\text{𝖳}}={𝑈}_{𝑖𝑖}$, ${𝑆}_{𝑖𝑗}^{\text{𝖳}}={𝑆}_{𝑖𝑗}$, and ${𝐾}_{𝑖𝑗}^{\text{𝖳}}=-{𝐾}_{𝑖𝑗}$,

So the matrix of $𝑇$ in this basis is diagonal, with $𝛼+𝛽$ appearing $𝑛+$$𝑛(𝑛-1)/2=$$𝑛(𝑛+1)/2$ times and $𝛼-𝛽$ appearing $𝑛(𝑛-1)/2$ times. By Definition 5.4.18 and Proposition 5.3.15, $det𝑇$ is the product of these diagonal entries, as claimed.

(b) The transpose operator is the case $𝛼=0$, $𝛽=1$, and the formula gives $1\cdot {(-1)}^{𝑛(𝑛-1)/2}$, which is the answer of Exercise 5.4.40.

(c) For $𝑛=2$, $𝑇({𝑈}_{11})=$$(𝛼+𝛽){𝑈}_{11}$, $𝑇({𝑈}_{22})=$$(𝛼+𝛽){𝑈}_{22}$, $𝑇({𝑈}_{12})=$$𝛼{𝑈}_{12}+$$𝛽{𝑈}_{21}$, and $𝑇({𝑈}_{21})=$$𝛽{𝑈}_{12}+$$𝛼{𝑈}_{21}$. In the basis ${𝑈}_{11},$${𝑈}_{12},$${𝑈}_{21},$${𝑈}_{22}$ the matrix of $𝑇$ is

which is block diagonal, so $det𝑇=$${(𝛼+𝛽)}^2({𝛼}^2-{𝛽}^2)=$${(𝛼+𝛽)}^3(𝛼-𝛽)$ (Corollary 5.4.27). This is the formula of (a) for $𝑛=2$, and the computation used nothing about $1+1$. (Over ${𝔽}_2$ it reads ${(𝛼+𝛽)}^4$.)

(d) By Proposition 5.4.20, $𝑇$ is invertible exactly when $det𝑇\ne 0$, that is, when $𝛼+𝛽\ne 0$ and, if $𝑛\ge 2$, also $𝛼-𝛽\ne 0$. If $𝛼=𝛽\ne 0$, then in the basis of (a) the diagonal entries are $1+1$ times $𝛼$ for ${𝑈}_{𝑖𝑖}$ and ${𝑆}_{𝑖𝑗}$, which is nonzero, and $0$ for ${𝐾}_{𝑖𝑗}$. So $ker𝑇=$$span({𝐾}_{𝑖𝑗}|𝑖<𝑗)$, the space of skew-symmetric matrices, of dimension $𝑛(𝑛-1)/2$. Directly: $𝛼(𝑋+{𝑋}^{\text{𝖳}})=0$ exactly when ${𝑋}^{\text{𝖳}}=-𝑋$.

(a) The $(𝑙,𝑗)$ entry of $𝑉$ is ${𝑥}_{𝑙}^{𝑗-1}$ (Definition 3.4.22), so the $(𝑖,𝑙)$ entry of ${𝑉}^{\text{𝖳}}$ is ${𝑥}_{𝑙}^{𝑖-1}$, and

So $𝐻={𝑉}^{\text{𝖳}}𝑉$, and by Theorem 5.4.10, Theorem 5.4.14, and Theorem 5.5.21,

(b) For $𝑛=2$, $𝐻=$$\left(\begin{array}{cc}2& {𝑥}_1+{𝑥}_2\\ {𝑥}_1+{𝑥}_2& {𝑥}_1^2+{𝑥}_2^2\end{array}\right)$, and