Problems for Chapter 4

(a) Let $𝑉$ be finite-dimensional and $𝑇\in \mathrm{L}(𝑉,𝑊)$. Then $im𝑇$ is finite-dimensional, and $dimker𝑇+$$dimim𝑇=$$dim𝑉$.

(b) Since $ker𝑇$ is a subspace of the finite-dimensional space $𝑉$, it is finite-dimensional (Proposition 4.1.21(a)). Let ${𝐮}_1,\dots ,{𝐮}_{𝑘}$ be a basis of $ker𝑇$ and extend it to a basis ${𝐮}_1,$$\dots ,$${𝐮}_{𝑘},$${𝐰}_1,$$\dots ,$${𝐰}_{𝑟}$ of $𝑉$ (Theorem 4.1.16), so $dim𝑉=𝑘+𝑟$. We show that $𝑇({𝐰}_1),$$\dots ,$$𝑇({𝐰}_{𝑟})$ is a basis of $im𝑇$; then $im𝑇$ is finite-dimensional with $dimim𝑇=$$𝑟=$$dim𝑉-$$dimker𝑇$.

Spanning. By Proposition 2.1.16, $im𝑇$ is spanned by the values of $𝑇$ on the basis, $\mathrm{𝟎},$$\dots ,$$\mathrm{𝟎},$$𝑇({𝐰}_1),$$\dots ,$$𝑇({𝐰}_{𝑟})$, and the zeros can be dropped.

Independence. If ${\sum }_{𝑗}{𝑏}_{𝑗}𝑇({𝐰}_{𝑗})=$$\mathrm{𝟎}$, then $𝑇\left({\sum }_{𝑗}{𝑏}_{𝑗}{𝐰}_{𝑗}\right)=$$\mathrm{𝟎}$, so ${\sum }_{𝑗}{𝑏}_{𝑗}{𝐰}_{𝑗}\in ker𝑇$ and ${\sum }_{𝑗}{𝑏}_{𝑗}{𝐰}_{𝑗}=$${\sum }_{𝑖}{𝑎}_{𝑖}{𝐮}_{𝑖}$ for some scalars ${𝑎}_{𝑖}$. The relation ${\sum }_{𝑖}{𝑎}_{𝑖}{𝐮}_{𝑖}-$${\sum }_{𝑗}{𝑏}_{𝑗}{𝐰}_{𝑗}=$$\mathrm{𝟎}$ among the vectors of a basis forces every ${𝑏}_{𝑗}=0$ (Theorem 1.5.5).

(c) Let ${𝐮}_1,\dots ,{𝐮}_{𝑘}$ be a basis of $ker𝑇$ and ${𝐯}_1,\dots ,{𝐯}_{𝑟}$ as described; here $𝑉$ is any vector space, and we only assume that $ker𝑇$ and $im𝑇$ are finite-dimensional. Let $\mathrm{ℒ︀}=$${𝐮}_1,$$\dots ,$${𝐮}_{𝑘},$${𝐯}_1,$$\dots ,$${𝐯}_{𝑟}$.

$\mathrm{ℒ︀}$ spans $𝑉$. Let $𝐯\in 𝑉$. Then $𝑇(𝐯)\in im𝑇$, so $𝑇(𝐯)=$${\sum }_{𝑗}{𝑐}_{𝑗}𝑇({𝐯}_{𝑗})$ for some scalars ${𝑐}_{𝑗}$, and $𝑇\left(𝐯-{\sum }_{𝑗}{𝑐}_{𝑗}{𝐯}_{𝑗}\right)=$$\mathrm{𝟎}$. So $𝐯-$${\sum }_{𝑗}{𝑐}_{𝑗}{𝐯}_{𝑗}\in$$ker𝑇$ is a combination of the ${𝐮}_{𝑖}$, and $𝐯$ is a combination of $\mathrm{ℒ︀}$.

$\mathrm{ℒ︀}$ is independent. If ${\sum }_{𝑖}{𝑎}_{𝑖}{𝐮}_{𝑖}+$${\sum }_{𝑗}{𝑏}_{𝑗}{𝐯}_{𝑗}=$$\mathrm{𝟎}$, apply $𝑇$: the first sum is killed, and ${\sum }_{𝑗}{𝑏}_{𝑗}𝑇({𝐯}_{𝑗})=$$\mathrm{𝟎}$ (Proposition 2.1.4). The $𝑇({𝐯}_{𝑗})$ are independent, so every ${𝑏}_{𝑗}=0$, and then ${\sum }_{𝑖}{𝑎}_{𝑖}{𝐮}_{𝑖}=\mathrm{𝟎}$ forces every ${𝑎}_{𝑖}=0$.

So $\mathrm{ℒ︀}$ is a basis of $𝑉$, $𝑉$ is finite-dimensional, and $dim𝑉=$$𝑘+$$𝑟=$$dimker𝑇+$$dimim𝑇$, which is rank–nullity again. The proof in (b) needs $dim𝑉<\infty$ before it starts, to extend a basis of the kernel to a basis of $𝑉$. The proof in (c) only needs finite spanning lists for the kernel and the image, and it produces one for $𝑉$. Contrapositively, if $𝑉$ is infinite-dimensional, then $ker𝑇$ or $im𝑇$ is. For instance multiplication by $𝑥$ on $𝐹[𝑥]$ has kernel $\{\mathrm{𝟎}\}$, so its image $𝑥𝐹[𝑥]$ is infinite-dimensional, and differentiation over $\mathbb{Q}$ has a $1$-dimensional kernel and the infinite-dimensional image $\mathbb{Q}[𝑥]$.

(a) For every $𝐴\in {𝐹}^{𝑚\times 𝑛}$, $rank{𝐴}^{\text{𝖳}}=rank𝐴$: the rows of $𝐴$ span a subspace of ${𝐹}^{𝑛}$ of the same dimension as the subspace of ${𝐹}^{𝑚}$ spanned by its columns.

(b) Let $𝑟=rank𝐴$. If $𝑟=0$, then $𝐴=0$ and ${𝐴}^{\text{𝖳}}=0$. Otherwise, by Theorem 4.3.21, $𝐴=𝐶𝑅$ with $𝐶\in {𝐹}^{𝑚\times 𝑟}$ and $𝑅\in {𝐹}^{𝑟\times 𝑛}$. By Proposition 2.4.23, ${𝐴}^{\text{𝖳}}={𝑅}^{\text{𝖳}}{𝐶}^{\text{𝖳}}$ with inner size $𝑟$, so $rank{𝐴}^{\text{𝖳}}\le$$𝑟=$$rank𝐴$ by Proposition 4.3.23(a). Applying this to ${𝐴}^{\text{𝖳}}$, whose transpose is $𝐴$, gives $rank𝐴\le rank{𝐴}^{\text{𝖳}}$.

(c) Let $𝐶\in {𝐹}^{𝑚\times 𝑟}$ consist of the columns of $𝐴$ with indices in $𝐽$. They are $𝑟$ independent vectors in the $𝑟$-dimensional space $im𝐴$ (Proposition 2.6.18), so they form a basis of $im𝐴$ (Theorem 4.1.10(b)). Every column of $𝐴$ is therefore a combination of the columns of $𝐶$, so $𝐴=𝐶𝑋$ for some $𝑋\in {𝐹}^{𝑟\times 𝑛}$ (Proposition 2.4.8). Taking the rows with indices in $𝐼$ on both sides, ${𝐴}_{𝐼\cdot }={𝐶}_{𝐼\cdot }𝑋$, where ${𝐴}_{𝐼\cdot }\in {𝐹}^{𝑟\times 𝑛}$ consists of the rows of $𝐴$ in $𝐼$ and ${𝐶}_{𝐼\cdot }={𝐴}_{𝐼𝐽}$. The $𝑟$ rows of ${𝐴}_{𝐼\cdot }$ are independent, so $rank{𝐴}_{𝐼\cdot }=𝑟$ by (a). By Theorem 4.3.3,

so the $𝑟\times 𝑟$ matrix ${𝐴}_{𝐼𝐽}$ has rank $𝑟$ and is invertible (Theorem 3.5.14).

The largest invertible submatrix. By (a) there are $𝑟$ independent rows and $𝑟$ independent columns, so by what we just proved, $𝐴$ has an invertible $𝑟\times 𝑟$ submatrix. Conversely, let ${𝐴}_{𝐼𝐽}$ be an invertible $𝑘\times 𝑘$ submatrix. A relation among the columns of $𝐴$ with indices in $𝐽$ gives, in the rows with indices in $𝐼$, the same relation among the columns of ${𝐴}_{𝐼𝐽}$, which are independent. So the $𝑘$ columns of $𝐴$ in $𝐽$ are independent, and $𝑘\le rank𝐴$.

The point of (c) is that the rows and the columns can be chosen separately: any $𝑟$ independent rows meet any $𝑟$ independent columns in an invertible submatrix. In Section 5.5 the same characterization of the rank is phrased with determinants.

(a) Proposition 4.1.26: let $𝑇\in \mathrm{L}(𝑉)$, $𝐯\in 𝑉$, and $𝑘\ge 0$ with ${𝑇}^{𝑘}𝐯\ne \mathrm{𝟎}$ and ${𝑇}^{𝑘+1}𝐯=\mathrm{𝟎}$. Then $𝐯,𝑇𝐯,\dots ,{𝑇}^{𝑘}𝐯$ is linearly independent. Corollary 4.1.27: if $dim𝑉=𝑛$ and ${𝑇}^{𝑚}=0$ for some $𝑚\ge 1$, then ${𝑇}^{𝑛}=0$.

(b) The proposition. For $𝑙\ge 𝑘+1$, ${𝑇}^{𝑙}𝐯=$${𝑇}^{𝑙-𝑘-1}({𝑇}^{𝑘+1}𝐯)=$$\mathrm{𝟎}$. Let ${\sum }_{𝑖=0}^{𝑘}{𝑎}_{𝑖}{𝑇}^{𝑖}𝐯=$$\mathrm{𝟎}$. We show ${𝑎}_{𝑖}=0$ by induction on $𝑖$. Suppose ${𝑎}_0=\cdots ={𝑎}_{𝑖-1}=0$ (no assumption for $𝑖=0$). Applying ${𝑇}^{𝑘-𝑖}$ to ${\sum }_{𝑙=𝑖}^{𝑘}{𝑎}_{𝑙}{𝑇}^{𝑙}𝐯=$$\mathrm{𝟎}$ gives

and every exponent in the sum is at least $𝑘+1$, so ${𝑎}_{𝑖}{𝑇}^{𝑘}𝐯=\mathrm{𝟎}$. Since ${𝑇}^{𝑘}𝐯\ne \mathrm{𝟎}$, ${𝑎}_{𝑖}=0$ (Proposition 1.2.35). So the list is independent (Theorem 1.5.5).

The corollary. Let $𝐯\ne \mathrm{𝟎}$. Since ${𝑇}^{𝑚}𝐯=\mathrm{𝟎}$, there is a largest $𝑘$ with ${𝑇}^{𝑘}𝐯\ne \mathrm{𝟎}$, and ${𝑇}^{𝑘+1}𝐯=\mathrm{𝟎}$. By the proposition, $𝐯,\dots ,{𝑇}^{𝑘}𝐯$ is an independent list of length $𝑘+1$ in $𝑉$, so $𝑘+1\le 𝑛$ (Theorem 4.1.10(a)), and ${𝑇}^{𝑛}𝐯=$${𝑇}^{𝑛-𝑘-1}({𝑇}^{𝑘+1}𝐯)=$$\mathrm{𝟎}$. Also ${𝑇}^{𝑛}\mathrm{𝟎}=\mathrm{𝟎}$. So ${𝑇}^{𝑛}=0$.

(c) Fix $𝑗$ with $1\le 𝑗\le 𝑘+1$. For $𝑖\ge 𝑘-𝑗+1$, ${𝑇}^{𝑗}({𝑇}^{𝑖}𝐯)=$${𝑇}^{𝑖+𝑗}𝐯=$$\mathrm{𝟎}$, since $𝑖+𝑗\ge 𝑘+1$. So the right side is contained in $ker{𝑇}^{𝑗}\cap 𝑊$. Conversely, let $𝐰=$${\sum }_{𝑖=0}^{𝑘}{𝑎}_{𝑖}{𝑇}^{𝑖}𝐯\in$$𝑊$ with ${𝑇}^{𝑗}𝐰=\mathrm{𝟎}$. The terms with $𝑖+𝑗\ge 𝑘+1$ vanish, so

The vectors ${𝑇}^{𝑗}𝐯,\dots ,{𝑇}^{𝑘}𝐯$ are entries of the independent chain, so ${𝑎}_0=\cdots ={𝑎}_{𝑘-𝑗}=0$, and $𝐰\in$$span({𝑇}^{𝑘-𝑗+1}𝐯,\dots ,{𝑇}^{𝑘}𝐯)$. These $𝑗$ vectors are independent, being entries of the chain, so $dim(ker{𝑇}^{𝑗}\cap 𝑊)=$$𝑗$.

So on $𝑊$ the kernels of the powers grow by exactly one dimension at each step, until $ker{𝑇}^{𝑘+1}\cap 𝑊=𝑊$. In particular $ker𝑇\cap 𝑊$ is the line spanned by the last vector ${𝑇}^{𝑘}𝐯$, the vector that becomes an eigenvector when the chain is shifted in Proposition 7.2.37.

If $𝑟=0$ or $𝑟=𝑛$, then $𝑃=0$ or $𝑃={𝐼}_{𝑛}$ (an idempotent of full rank is invertible, and ${𝑃}^2=𝑃$ then gives $𝑃={𝐼}_{𝑛}$), so $𝛿=0$, and all three statements hold with $ker𝛿={𝐹}^{𝑛\times 𝑛}$ and $im𝛿=\{0\}$. So let $1\le 𝑟\le 𝑛-1$. By Theorem 4.4.27, ${𝐹}^{𝑛}=im𝑃\oplus ker𝑃$, and $𝑃$ fixes $im𝑃$ pointwise and kills $ker𝑃$. Let $𝑆\in {𝐹}^{𝑛\times 𝑛}$ have as columns a basis of $im𝑃$ followed by a basis of $ker𝑃$; together they form a basis of ${𝐹}^{𝑛}$ (Theorem 4.4.13), so $𝑆$ is invertible. In this basis ${𝑇}_{𝑃}$ has the matrix $𝐷=\left(\begin{array}{cc}{𝐼}_{𝑟}& 0\\ 0& 0\end{array}\right)$, so ${𝑆}^{-1}𝑃𝑆=𝐷$ (Theorem 4.5.17). For $𝑋\in {𝐹}^{𝑛\times 𝑛}$ write $𝑌={𝑆}^{-1}𝑋𝑆$. Then

The map $𝑋\mapsto {𝑆}^{-1}𝑋𝑆$ is linear and invertible, with inverse $𝑌\mapsto 𝑆𝑌{𝑆}^{-1}$, so it carries $ker𝛿$ onto $ker{𝛿}_{𝐷}$ and $im𝛿$ onto $im{𝛿}_{𝐷}$, where ${𝛿}_{𝐷}(𝑌)=$$𝐷𝑌-$$𝑌𝐷$ (Proposition 4.5.5). In blocks, with ${𝑌}_{11}\in {𝐹}^{𝑟\times 𝑟}$, Theorem 2.4.34 gives

(a) ${𝛿}_{𝐷}(𝑌)=0$ if and only if ${𝑌}_{12}=0$ and ${𝑌}_{21}=0$. So $ker{𝛿}_{𝐷}$ is the space of block diagonal matrices, of dimension ${𝑟}^2+{(𝑛-𝑟)}^2$, and $ker𝛿$ has the same dimension.

(b) By rank–nullity, $rank𝛿=$${𝑛}^2-$${𝑟}^2-$${(𝑛-𝑟)}^2=$$2𝑟(𝑛-𝑟)$. The image of ${𝛿}_{𝐷}$ consists of the block off-diagonal matrices $\left(\begin{array}{cc}0& {𝑍}_{12}\\ {𝑍}_{21}& 0\end{array}\right)$, since ${𝑌}_{12}$ and $-{𝑌}_{21}$ can be anything. For ${𝑍}^{\prime }=$$\left(\begin{array}{cc}{𝑍}_{11}& {𝑍}_{12}\\ {𝑍}_{21}& {𝑍}_{22}\end{array}\right)$,

so ${𝑍}^{\prime }\in im{𝛿}_{𝐷}$ if and only if $𝐷{𝑍}^{\prime }𝐷=0$ and $({𝐼}_{𝑛}-𝐷){𝑍}^{\prime }({𝐼}_{𝑛}-𝐷)=$$0$. For $𝑍=𝑆{𝑍}^{\prime }{𝑆}^{-1}$, we have $𝑃𝑍𝑃=$$𝑆(𝐷{𝑍}^{\prime }𝐷){𝑆}^{-1}$ and $({𝐼}_{𝑛}-𝑃)𝑍({𝐼}_{𝑛}-𝑃)=$$𝑆(({𝐼}_{𝑛}-𝐷){𝑍}^{\prime }({𝐼}_{𝑛}-𝐷)){𝑆}^{-1}$, using ${𝐼}_{𝑛}-$$𝑃=$$𝑆({𝐼}_{𝑛}-𝐷){𝑆}^{-1}$. So $𝑍\in im𝛿$ if and only if $𝑃𝑍𝑃=0$ and $({𝐼}_{𝑛}-𝑃)𝑍({𝐼}_{𝑛}-𝑃)=$$0$.

(c) By (a) and (b), $dimker𝛿+$$dimim𝛿=$${𝑛}^2$. A matrix in both is, after conjugation by $𝑆$, both block diagonal and block off-diagonal, hence $0$. So ${𝐹}^{𝑛\times 𝑛}=$$ker𝛿\oplus$$im𝛿$ by Proposition 4.4.21. Let ${𝑋}_0=$$𝑃𝑋𝑃+$$({𝐼}_{𝑛}-𝑃)𝑋({𝐼}_{𝑛}-𝑃)$. It commutes with $𝑃$: since ${𝑃}^2=𝑃$ and $𝑃({𝐼}_{𝑛}-𝑃)=$$0=$$({𝐼}_{𝑛}-𝑃)𝑃$, both $𝑃{𝑋}_0$ and ${𝑋}_0𝑃$ equal $𝑃𝑋𝑃$. And $𝑋-$${𝑋}_0=$$𝑃𝑋({𝐼}_{𝑛}-𝑃)+$$({𝐼}_{𝑛}-𝑃)𝑋𝑃$ satisfies the two conditions of (b), because $𝑃(𝑋-{𝑋}_0)𝑃=0$ and $({𝐼}_{𝑛}-𝑃)(𝑋-{𝑋}_0)({𝐼}_{𝑛}-𝑃)=$$0$ by the same two identities. So $𝑋={𝑋}_0+(𝑋-{𝑋}_0)$ is the decomposition.

Choosing the basis made the computation a matter of reading blocks. The formula in (c) needs no basis at all, and could have been guessed from the blocks: $𝑃𝑋𝑃$ and $({𝐼}_{𝑛}-𝑃)𝑋({𝐼}_{𝑛}-𝑃)$ are the two diagonal blocks of $𝑋$ seen through the decomposition ${𝐹}^{𝑛}=im𝑃\oplus ker𝑃$.

(a) All the maps involved send ${𝐹[𝑥]}_{\le 𝑛}$ into itself, since $𝑝(𝑥+𝑗)$ has the same degree as $𝑝$. By the binomial theorem, for $𝑗\ge 1$ the polynomial $\mathrm{\Delta }({𝑥}^{𝑗})=$${(𝑥+1)}^{𝑗}-$${𝑥}^{𝑗}$ has degree at most $𝑗-1$, and its coefficient of ${𝑥}^{𝑗-1}$ is $𝑗\ast 1$; and $\mathrm{\Delta }(1)=\mathrm{𝟎}$. Let $𝑞=$${𝑞}_0+$${𝑞}_1𝑥+$$\cdots +$${𝑞}_{𝑑}{𝑥}^{𝑑}$ with $𝑑\ge 1$ and $𝑐={𝑞}_{𝑑}\ne 0$. By linearity, $\mathrm{\Delta }𝑞=$${\sum }_{𝑗=0}^{𝑑}{𝑞}_{𝑗}\mathrm{\Delta }({𝑥}^{𝑗})$, and only the term $𝑗=𝑑$ can contribute to the coefficient of ${𝑥}^{𝑑-1}$, which is therefore $𝑑𝑐$. If $𝑑𝑐\ne 0$, then $\mathrm{\Delta }𝑞$ has degree exactly $𝑑-1$ and leading coefficient $𝑑𝑐$; and $\mathrm{\Delta }$ kills constants. Let $𝑝$ have degree $𝑛$ and leading coefficient $𝑐$. By induction on $𝑘$, for $𝑘\le 𝑛$, ${\mathrm{\Delta }}^{𝑘}𝑝$ has degree $𝑛-𝑘$ and leading coefficient $𝑛(𝑛-1)\cdots (𝑛-𝑘+1)𝑐$. This coefficient is nonzero: its integer factors $𝑛,𝑛-1,\dots ,𝑛-𝑘+1$ are factors of $𝑛!$, which is nonzero in $𝐹$, so none of them is $0$ in $𝐹$, and a product of nonzero elements of a field is nonzero. In particular ${\mathrm{\Delta }}^{𝑛}𝑝=𝑛!𝑐$ is a nonzero constant, and ${\mathrm{\Delta }}^{𝑛+1}𝑝=\mathrm{𝟎}$.

By Proposition 4.1.26, applied to $\mathrm{\Delta }$ on ${𝐹[𝑥]}_{\le 𝑛}$, the list $𝑝,\mathrm{\Delta }𝑝,\dots ,{\mathrm{\Delta }}^{𝑛}𝑝$ is independent. It has $𝑛+1=dim{𝐹[𝑥]}_{\le 𝑛}$ entries, so it is a basis (Theorem 4.1.10(b)). Since $𝑆$ commutes with $\text{id}$, the binomial theorem for commuting operators (which has the same proof as for numbers, using Proposition 2.1.24) gives

So every ${\mathrm{\Delta }}^{𝑘}𝑝$ with $𝑘\le 𝑛$ lies in the span of $𝑝(𝑥),$$𝑝(𝑥+1),$$\dots ,$$𝑝(𝑥+𝑛)$, and this span is all of ${𝐹[𝑥]}_{\le 𝑛}$. A spanning list of length $𝑛+1=dim{𝐹[𝑥]}_{\le 𝑛}$ is a basis (Corollary 4.1.11(b)).

(b) For $𝑝={𝑥}^2$, $\mathrm{\Delta }𝑝=2𝑥+1$ and ${\mathrm{\Delta }}^2𝑝=2$. By the formula above, $2=$${\mathrm{\Delta }}^2𝑝=$${(𝑥+2)}^2-$$2{(𝑥+1)}^2+$${𝑥}^2$, so

Check: $\frac{1}{2}({𝑥}^2+4𝑥+4)-$$({𝑥}^2+2𝑥+1)+$$\frac{1}{2}{𝑥}^2=$$2-$$1=$$1$.

(c) Over ${𝔽}_{𝑝}$, the binomial coefficients $\left(\genfrac{}{}{0ex}{}{𝑝}{𝑖}\right)$ with $0<𝑖<𝑝$ are $0$, so $𝑆({𝑥}^{𝑝})=$${(𝑥+1)}^{𝑝}=$${𝑥}^{𝑝}+$$1$, and $𝑆(1)=1$. By induction on $𝑗$, ${𝑆}^{𝑗}({𝑥}^{𝑝})=$${𝑥}^{𝑝}+$$𝑗\ast 1$. So the translates of ${𝑥}^{𝑝}$ are the polynomials ${𝑥}^{𝑝}+𝑐$ with $𝑐\in {𝔽}_{𝑝}$; they lie in $span({𝑥}^{𝑝},1)$ and include ${𝑥}^{𝑝}$ and ${𝑥}^{𝑝}+1$, so they span exactly this $2$-dimensional space. For $𝑛=𝑝$ the factor $𝑝$ of $𝑛!$ is $0$ in ${𝔽}_{𝑝}$, the chain ${𝑥}^{𝑝},$$\mathrm{\Delta }{𝑥}^{𝑝}=$$1,$${\mathrm{\Delta }}^2{𝑥}^{𝑝}=$$\mathrm{𝟎}$ stops after two steps, and the argument of (a) breaks exactly there.

(a) A combination of the rows of $𝑀$ with coefficients ${𝑎}_{𝑖}$ and ${𝑏}_{𝑗}$ is

with $𝐮={\sum }_{𝑖}{𝑎}_{𝑖}{𝐮}_{𝑖}\in 𝑈$ and $𝐰={\sum }_{𝑗}{𝑏}_{𝑗}{𝐰}_{𝑗}\in 𝑊$, and every such pair arises. So the row space is $𝑍$. The map $𝜋(𝐚,𝐛)=𝐚$ is linear, and $𝜋(𝑍)=$$\{𝐮+𝐰\}=$$𝑈+$$𝑊$. A pair $(𝐮+𝐰,𝐮)$ lies in $ker𝜋$ exactly when $𝐮=-𝐰$, that is, $𝐮\in 𝑈\cap 𝑊$; and for $𝐮\in 𝑈\cap 𝑊$, $(\mathrm{𝟎},𝐮)=$$(𝐮+(-𝐮),𝐮)\in$$𝑍$. So $ker(𝜋{\upharpoonright }_{𝑍})=$$\{(\mathrm{𝟎},𝐮)|𝐮\in 𝑈\cap 𝑊\}$, which is isomorphic to $𝑈\cap 𝑊$ by $(\mathrm{𝟎},𝐮)\mapsto 𝐮$.

(b) Row operations do not change the row space, and the nonzero rows of $𝑅$ form a basis of it (Proposition 4.3.13(b)), so they form a basis of $𝑍$. In $𝑅$ the pivots move to the right from one row to the next (Theorem 3.3.12), so the nonzero rows split into $𝑠$ rows ${𝐫}_1,\dots ,{𝐫}_{𝑠}$ with their pivot in the left half, followed by $𝑑$ rows of the form $(\mathrm{𝟎},{𝐜}_1),$$\dots ,$$(\mathrm{𝟎},{𝐜}_{𝑑})$.

The left halves of the ${𝐫}_{𝑖}$ form a basis of $𝑈+𝑊$. They span $𝜋(𝑍)=𝑈+𝑊$, since $𝜋$ of the other basis vectors of $𝑍$ is $\mathrm{𝟎}$. They are independent: in a relation ${\sum }_{𝑖}{𝑐}_{𝑖}𝜋({𝐫}_{𝑖})=$$\mathrm{𝟎}$, look at the pivot column of ${𝐫}_{𝑖}$, which lies in the left half and in which every other row of $𝑅$ has a $0$; it gives ${𝑐}_{𝑖}=0$.

The ${𝐜}_{𝑗}$ form a basis of $𝑈\cap 𝑊$. The vectors $(\mathrm{𝟎},{𝐜}_{𝑗})$ are independent, being nonzero rows of $𝑅$, and they lie in $ker(𝜋{\upharpoonright }_{𝑍})$. By rank–nullity for $𝜋{\upharpoonright }_{𝑍}$, this kernel has dimension $dim𝑍-$$dim(𝑈+𝑊)=$$(𝑠+𝑑)-$$𝑠=$$𝑑$. So they form a basis of it (Theorem 4.1.10(b)), and by (a) the ${𝐜}_{𝑗}$ form a basis of $𝑈\cap 𝑊$. In passing, $dim(𝑈+𝑊)+$$dim(𝑈\cap 𝑊)=$$𝑠+$$𝑑=$$dim𝑍=$$dim𝑈+$$dim𝑊$ when the ${𝐮}_{𝑖}$ and the ${𝐰}_{𝑗}$ are bases, which is Theorem 4.4.19 once more.

(c) The matrix is

Over ${𝔽}_2$, where $-1=1$, Gauss–Jordan elimination gives

So $𝑈+𝑊$ has basis $(1,0,0,1),$$(0,1,0,1),$$(0,0,1,1)$, the hyperplane ${𝑥}_1+{𝑥}_2+{𝑥}_3+{𝑥}_4=0$, and $𝑈\cap$$𝑊=$$span((1,0,1,0))$. Indeed $(1,0,1,0)=$${𝐮}_1+$${𝐮}_2=$${𝐰}_1+$${𝐰}_2$ over ${𝔽}_2$.

Over $\mathbb{Q}$, the same elimination ends with a pivot in each of the first four columns, since

so $𝑈+𝑊={\mathbb{Q}}^4$ and $𝑈\cap 𝑊=\{\mathrm{𝟎}\}$. Directly: $𝑈=$$\{(𝑎,𝑎+𝑏,𝑏,0)\}$ and $𝑊=$$\{(𝑑,0,𝑐,𝑐-𝑑)\}$, and a common vector has $𝑎=𝑏=𝑐=𝑑$ and $2𝑎=0$. Over $\mathbb{Q}$ this forces $𝑎=0$, and over ${𝔽}_2$ it does not.

The column method of Example 4.4.20 finds the intersection as a relation among the given vectors; this method finds bases of both spaces in one reduction, with no relation to interpret.

(a) Let $𝐺={𝐼}_{𝑛}-𝐴-𝐵$. Using ${𝐴}^2=𝐴$ and ${𝐵}^2=𝐵$,

So $𝐴𝐺=𝐺𝐵$, and since $𝐺$ is invertible, $𝐵={𝐺}^{-1}𝐴𝐺$. So $𝐴\sim 𝐵$, and $rank𝐴=rank𝐵$ (Proposition 4.5.27(a)). The rank statement alone also follows from Proposition 4.3.1: $rank𝐴=$$rank(𝐴𝐺)=$$rank(𝐺𝐵)=$$rank𝐵$.

(b) $𝐴={𝐼}_{𝑛}$ and $𝐵=0$ are idempotent of ranks $𝑛$ and $0$, and ${𝐼}_{𝑛}-𝐴-𝐵=0$ is not invertible.

(c) ${(𝐴+𝐵)}^2=$${𝐴}^2+$$𝐴𝐵+$$𝐵𝐴+$${𝐵}^2=$$𝐴+$$𝐵$ (Proposition 2.4.14). For every $𝐱$, $(𝐴+𝐵)𝐴𝐱=$${𝐴}^2𝐱+$$𝐵𝐴𝐱=$$𝐴𝐱$, so $im𝐴\subseteq im(𝐴+𝐵)$; similarly $(𝐴+𝐵)𝐵𝐱=𝐵𝐱$ gives $im𝐵\subseteq im(𝐴+𝐵)$. Hence $im𝐴+$$im𝐵\subseteq$$im(𝐴+𝐵)$ (Proposition 4.4.2), and the converse containment holds because $(𝐴+𝐵)𝐱=$$𝐴𝐱+$$𝐵𝐱$. So $im(𝐴+𝐵)=$$im𝐴+$$im𝐵$. If $𝐲\in im𝐴\cap im𝐵$, say $𝐲=𝐴𝐱=𝐵𝐳$, then $𝐲=$${𝐴}^2𝐱=$$𝐴𝐲=$$𝐴𝐵𝐳=$$\mathrm{𝟎}$. By Theorem 4.4.19,

Over $\mathbb{Q}$, $ℝ$, or $ℂ$, (c) also follows from $tr=rank$ for idempotents (Exercise 4.3.43), since the trace is additive. The argument above works over every field, where the trace of an idempotent only determines its rank modulo the characteristic.

(a) Let $𝐴$ be the matrix of $𝑁$ in some basis, so ${𝐴}^{𝑘}=0$ and $rank𝐴=rank𝑁$ (Proposition 4.2.15). By Exercise 4.3.30(c) with all $𝑘$ factors equal to $𝐴$,

so $rank𝑁\le$$(𝑘-1)𝑛/𝑘$ and $dimker𝑁=$$𝑛-$$rank𝑁\ge$$𝑛/𝑘$. The kernel proof of Exercise 4.2.52 and this one bound the same quantity; the kernel proof is the kernel proof of Sylvester’s inequality (Example 4.8.3), applied $𝑘$ times.

(b) Since ${𝑁}^{𝑘}=0$, ${\sum }_{𝑗=1}^{𝑘}{𝑑}_{𝑗}=$$dimker{𝑁}^{𝑘}-$$dimker{𝑁}^0=$$𝑛$. By Exercise 4.2.52(b), ${𝑑}_{𝑗}\le {𝑑}_1=𝑛/𝑘$ for every $𝑗$. A sum of $𝑘$ numbers, each at most $𝑛/𝑘$, equals $𝑛$ only if every term equals $𝑛/𝑘$. So $dimker{𝑁}^{𝑗}=$${𝑑}_1+$$\cdots +$${𝑑}_{𝑗}=$$𝑗𝑛/𝑘$ for $0\le 𝑗\le 𝑘$. In particular $dimker{𝑁}^{𝑘-1}=$$𝑛-$$𝑛/𝑘<$$𝑛$, so ${𝑁}^{𝑘-1}\ne 0$: a nilpotent operator whose kernel is as small as the bound allows takes the full $𝑘$ steps to die.