Problems for Chapter 2

(a) Theorem 2.6.11: a linear map $𝑇\in \mathrm{L}(𝑉,𝑊)$ is injective if and only if $ker𝑇=\{\mathrm{𝟎}\}$. Proposition 2.6.14: $𝑇\in \mathrm{L}(𝑉,𝑊)$ is injective if and only if $𝑇$ sends every linearly independent list ${𝐯}_1,\dots ,{𝐯}_{𝑘}$ in $𝑉$ to a linearly independent list $𝑇({𝐯}_1),$$\dots ,$$𝑇({𝐯}_{𝑘})$.

(b) The theorem. Suppose $𝑇$ is injective, and let $𝐯\in ker𝑇$. Then $𝑇(𝐯)=\mathrm{𝟎}=𝑇(\mathrm{𝟎})$ by Proposition 2.1.4, so $𝐯=\mathrm{𝟎}$. Conversely, suppose $ker𝑇=\{\mathrm{𝟎}\}$, and let $𝑇(𝐯)=𝑇({𝐯}^{\prime })$. By Proposition 2.1.4, $𝑇(𝐯-{𝐯}^{\prime })=$$𝑇(𝐯)-$$𝑇({𝐯}^{\prime })=$$\mathrm{𝟎}$, so $𝐯-{𝐯}^{\prime }\in ker𝑇$, that is, $𝐯={𝐯}^{\prime }$.

The proposition. Suppose $𝑇$ is injective, let ${𝐯}_1,\dots ,{𝐯}_{𝑘}$ be independent, and let ${\sum }_{𝑖}{𝑎}_{𝑖}𝑇({𝐯}_{𝑖})=$$\mathrm{𝟎}$. By linearity, $𝑇\left({\sum }_{𝑖}{𝑎}_{𝑖}{𝐯}_{𝑖}\right)=$$\mathrm{𝟎}$, so ${\sum }_{𝑖}{𝑎}_{𝑖}{𝐯}_{𝑖}\in$$ker𝑇=$$\{\mathrm{𝟎}\}$, and independence gives every ${𝑎}_{𝑖}=0$. So the image list is independent (Theorem 1.5.5). Conversely, suppose $𝑇$ is not injective. By the theorem there is $𝐯\ne \mathrm{𝟎}$ in $ker𝑇$. The list $𝐯$ is independent (Proposition 1.5.15(b)), but its image is the list $\mathrm{𝟎}$, which is dependent (Proposition 1.5.15(c)).

(c) Additive maps. The proof of the theorem used two facts: $𝑇(\mathrm{𝟎})=\mathrm{𝟎}$, and $𝑇(𝐯-{𝐯}^{\prime })=$$𝑇(𝐯)-$$𝑇({𝐯}^{\prime })$. Both follow from additivity alone. First, $𝑇(\mathrm{𝟎})=$$𝑇(\mathrm{𝟎}+\mathrm{𝟎})=$$𝑇(\mathrm{𝟎})+$$𝑇(\mathrm{𝟎})$, and adding $-𝑇(\mathrm{𝟎})$ gives $𝑇(\mathrm{𝟎})=\mathrm{𝟎}$. Next, $𝑇({𝐯}^{\prime })+$$𝑇(-{𝐯}^{\prime })=$$𝑇(\mathrm{𝟎})=$$\mathrm{𝟎}$, so $𝑇(-{𝐯}^{\prime })=-𝑇({𝐯}^{\prime })$, and then $𝑇(𝐯-{𝐯}^{\prime })=$$𝑇(𝐯)+$$𝑇(-{𝐯}^{\prime })=$$𝑇(𝐯)-$$𝑇({𝐯}^{\prime })$. With these two facts the proof runs word for word, so an additive map is injective if and only if it sends only $\mathrm{𝟎}$ to $\mathrm{𝟎}$.

Homogeneous maps. Define $𝑇:{𝔽}_2^2\to {𝔽}_2$ by $𝑇(\mathrm{𝟎})=0$ and $𝑇(𝐯)=1$ for $𝐯\ne \mathrm{𝟎}$. The only scalars are $0$ and $1$, and $𝑇(0𝐯)=$$𝑇(\mathrm{𝟎})=$$0=$$0\cdot$$𝑇(𝐯)$ and $𝑇(1𝐯)=$$𝑇(𝐯)=$$1\cdot$$𝑇(𝐯)$, so $𝑇$ is homogeneous. By construction $𝑇(𝐯)=0$ only for $𝐯=\mathrm{𝟎}$. But $𝑇({𝐞}_1)=𝑇({𝐞}_2)=1$ with ${𝐞}_1\ne {𝐞}_2$, so $𝑇$ is not injective. It is not additive either, since $𝑇({𝐞}_1)+$$𝑇({𝐞}_2)=$$0\ne$$1=$$𝑇({𝐞}_1+{𝐞}_2)$, and additivity is the half of linearity that the kernel test needs.

(a) Let $𝑈$, $𝑉$, $𝑊$ have bases $\mathrm{𝒜︀}={𝐮}_1,\dots ,{𝐮}_{𝑛}$, $\mathrm{ℬ︀}$, and $\mathrm{𝒞︀}$. For $𝑆\in \mathrm{L}(𝑈,𝑉)$ and $𝑇\in \mathrm{L}(𝑉,𝑊)$,

(b) By Definition 2.3.1, column $𝑗$ of ${[𝑇𝑆]}_{\mathrm{𝒞︀}\mathrm{𝒜︀}}$ is ${[𝑇(𝑆({𝐮}_{𝑗}))]}_{\mathrm{𝒞︀}}$. By Theorem 2.3.9, applied to $𝑇$ at the vector $𝑆({𝐮}_{𝑗})$, this is ${[𝑇]}_{\mathrm{𝒞︀}\mathrm{ℬ︀}}{[𝑆({𝐮}_{𝑗})]}_{\mathrm{ℬ︀}}$, and ${[𝑆({𝐮}_{𝑗})]}_{\mathrm{ℬ︀}}$ is column $𝑗$ of ${[𝑆]}_{\mathrm{ℬ︀}\mathrm{𝒜︀}}$. By Definition 2.4.1, ${[𝑇]}_{\mathrm{𝒞︀}\mathrm{ℬ︀}}$ times column $𝑗$ of ${[𝑆]}_{\mathrm{ℬ︀}\mathrm{𝒜︀}}$ is column $𝑗$ of ${[𝑇]}_{\mathrm{𝒞︀}\mathrm{ℬ︀}}{[𝑆]}_{\mathrm{ℬ︀}\mathrm{𝒜︀}}$. The two matrices have the same columns.

Associativity. Let $𝐴\in {𝐹}^{\ell \times 𝑚}$, $𝐵\in {𝐹}^{𝑚\times 𝑛}$, $𝐶\in {𝐹}^{𝑛\times 𝑝}$, and use standard bases throughout, so that ${[{𝑇}_{𝑋}]}_{\mathrm{ℰ︀}\mathrm{ℰ︀}}=𝑋$ for every matrix $𝑋$ (Proposition 2.3.16). Applying (a) twice,

Composition of functions is associative (Proposition 2.1.24(a)), so $({𝑇}_{𝐴}{𝑇}_{𝐵}){𝑇}_{𝐶}=$${𝑇}_{𝐴}({𝑇}_{𝐵}{𝑇}_{𝐶})$, and the two matrices are equal.

(c) If $\mathrm{ℬ︀}=\mathrm{𝒞︀}$, then (a) with $𝑆=𝑇$ and all three bases equal to $\mathrm{ℬ︀}$ gives ${[𝑇]}_{\mathrm{ℬ︀}\mathrm{ℬ︀}}{[𝑇]}_{\mathrm{ℬ︀}\mathrm{ℬ︀}}=$${[{𝑇}^2]}_{\mathrm{ℬ︀}\mathrm{ℬ︀}}$.

Conversely, suppose the identity holds for every $𝑇$, and take $𝑇={\text{id}}_{𝑉}$. With $𝑃={[{\text{id}}_{𝑉}]}_{\mathrm{𝒞︀}\mathrm{ℬ︀}}$ it reads ${𝑃}^2=$${[{\text{id}}_{𝑉}]}_{\mathrm{𝒞︀}\mathrm{ℬ︀}}=$$𝑃$. The matrix $𝑃$ is invertible, with inverse ${[{\text{id}}_{𝑉}]}_{\mathrm{ℬ︀}\mathrm{𝒞︀}}$ (Example 2.5.21), so ${𝑃}^2=𝑃$ forces $𝑃={𝐼}_{𝑛}$ (Exercise 2.5.32(c)). By Exercise 2.3.26, $\mathrm{ℬ︀}=\mathrm{𝒞︀}$.

In the product ${[𝑇]}_{\mathrm{𝒞︀}\mathrm{ℬ︀}}{[𝑇]}_{\mathrm{𝒞︀}\mathrm{ℬ︀}}$, the right factor produces $\mathrm{𝒞︀}$-coordinates and the left factor expects $\mathrm{ℬ︀}$-coordinates. Part (c) says that this mismatch is harmless only when the two bases coincide.

(a) Theorem 2.7.4: for $𝐴\in {𝐹}^{𝑚\times 𝑛}$ the following are equivalent. (I) ${𝑇}_{𝐴}$ is surjective. (II) For every $𝐲\in {𝐹}^{𝑚}$, $𝐴𝐱=𝐲$ has at least one solution. (III) The columns of $𝐴$ span ${𝐹}^{𝑚}$. (IV) $im{𝑇}_{𝐴}={𝐹}^{𝑚}$. (V) ${𝑇}_{𝐴}$ has a right inverse, a linear map $𝑆:{𝐹}^{𝑚}\to {𝐹}^{𝑛}$ with ${𝑇}_{𝐴}\circ 𝑆={\text{id}}_{{𝐹}^{𝑚}}$.

(b) Assume (II). For $𝑖=1,\dots ,𝑚$ choose ${𝐱}_{𝑖}\in {𝐹}^{𝑛}$ with $𝐴{𝐱}_{𝑖}={𝐞}_{𝑖}$, and let $𝐵\in {𝐹}^{𝑛\times 𝑚}$ have columns ${𝐱}_1,\dots ,{𝐱}_{𝑚}$. By the column view (Proposition 2.4.8(b)), column $𝑖$ of $𝐴𝐵$ is $𝐴{𝐱}_{𝑖}={𝐞}_{𝑖}$, so $𝐴𝐵={𝐼}_{𝑚}$. By Theorem 2.4.2(b), ${𝑇}_{𝐴}\circ$${𝑇}_{𝐵}=$${𝑇}_{𝐴𝐵}=$${\text{id}}_{{𝐹}^{𝑚}}$, so $𝑆={𝑇}_{𝐵}$ is a right inverse. In the language of Theorem 2.2.1, $𝑆$ is the linear map with $𝑆({𝐞}_{𝑖})={𝐱}_{𝑖}$, and ${𝑇}_{𝐴}\circ 𝑆$ agrees with $\text{id}$ on the standard basis.

The choices are made on a basis because only there do they automatically fit together. A choice of one solution for each $𝐲$ separately is a function ${𝐹}^{𝑚}\to {𝐹}^{𝑛}$, but usually not a linear one (Non-example 2.7.10(b)). Once ${𝐱}_1,\dots ,{𝐱}_{𝑚}$ are chosen, the extension sends $𝐲={\sum }_{𝑖}{𝑦}_{𝑖}{𝐞}_{𝑖}$ to ${\sum }_{𝑖}{𝑦}_{𝑖}{𝐱}_{𝑖}$, which is a solution of $𝐴𝐱=𝐲$ by linearity: no further choice is made.

(c) By the column view, $𝐶\in {𝔽}_{𝑝}^{𝑛\times 𝑚}$ satisfies $𝐴𝐶={𝐼}_{𝑚}$ if and only if its columns satisfy $𝐴{𝐜}_{𝑖}={𝐞}_{𝑖}$ for $𝑖=1,\dots ,𝑚$. The solutions of $𝐴𝐱={𝐞}_{𝑖}$ form the fiber ${𝑇}_{𝐴}^{-1}({𝐞}_{𝑖})$, which is nonempty since ${𝑇}_{𝐴}$ is surjective, and is a translate of $ker{𝑇}_{𝐴}$ (Corollary 2.6.31(b)); so it has $|ker{𝑇}_{𝐴}|$ elements. By Exercise 2.6.44(a),

so $|ker{𝑇}_{𝐴}|={𝑝}^{𝑛-𝑚}$. The $𝑚$ columns are chosen independently, so there are ${({𝑝}^{𝑛-𝑚})}^{𝑚}=$${𝑝}^{𝑚(𝑛-𝑚)}$ right inverses.

In Exercise 2.5.36(c), ${𝑇}_{{𝐴}^{\text{𝖳}}}$ is surjective for the matrix ${𝐴}^{\text{𝖳}}\in {𝔽}_{𝑝}^{2\times 3}$ there, and it has ${𝑝}^2$ right inverses; the formula gives ${𝑝}^{2(3-2)}={𝑝}^2$. A square surjective $𝐴$ has exactly one right inverse, as it must (Theorem 3.5.23).

(a) ${𝐿}_{𝐴,𝐵}$ is linear by Proposition 2.4.14(c) and (d). For every $𝑋$, associativity gives

and by Proposition 2.4.23 and Proposition 2.4.22,

(b) By Lemma 2.4.24(a), the coordinates of a matrix with respect to $\mathrm{𝒰︀}$ are its entries. So the diagonal entry of ${[\mathrm{\Phi }]}_{\mathrm{𝒰︀}\mathrm{𝒰︀}}$ in the position of ${𝑈}_{𝑖𝑗}$ is the $(𝑖,𝑗)$ entry of $\mathrm{\Phi }({𝑈}_{𝑖𝑗})$, whatever order $\mathrm{𝒰︀}$ is listed in, and $tr\mathrm{\Phi }$ is the sum of these entries over all $(𝑖,𝑗)$. For $\mathrm{\Phi }={𝐿}_{𝐴,𝐵}$, by Lemma 2.4.24(b),

and ${\sum }_{𝑖,𝑗}{𝑎}_{𝑖𝑖}{𝑏}_{𝑗𝑗}=$$(tr𝐴)(tr𝐵)$.

For $𝜏$, $𝜏({𝑈}_{𝑖𝑗})={𝑈}_{𝑗𝑖}$, whose $(𝑖,𝑗)$ entry is ${𝛿}_{𝑖𝑗}$. So $tr𝜏=$${\sum }_{𝑖,𝑗}{𝛿}_{𝑖𝑗}=$$𝑛\ast 1$.

For ${𝐶}_{𝐴}=$${𝐿}_{𝐴,{𝐼}_{𝑛}}-$${𝐿}_{{𝐼}_{𝑛},𝐴}$, Proposition 2.4.12 and Proposition 2.4.27 give $tr{𝐶}_{𝐴}=$$(tr𝐴)(𝑛\ast 1)-$$(𝑛\ast 1)(tr𝐴)=$$0$.

(c) For all indices, associativity and Lemma 2.4.24(b) give

So ${𝐿}_{{𝑈}_{𝑖𝑗},{𝑈}_{𝑘𝑙}}$ sends ${𝑈}_{𝑗𝑘}$ to ${𝑈}_{𝑖𝑙}$ and every other matrix unit to $0$. These are the maps of Exercise 2.2.34 for $𝑉=𝑊={𝐹}^{𝑛\times 𝑛}$ with the basis $\mathrm{𝒰︀}$ on both sides: as $(𝑖,𝑗,𝑘,𝑙)$ runs over all quadruples, the pair $({𝑈}_{𝑗𝑘},{𝑈}_{𝑖𝑙})$ runs over all pairs of matrix units exactly once. By that exercise they form a basis of $\mathrm{L}({𝐹}^{𝑛\times 𝑛})$. Every $\mathrm{\Phi }$ is therefore a combination $\sum {𝑐}_{𝑖𝑗𝑘𝑙}{𝐿}_{{𝑈}_{𝑖𝑗},{𝑈}_{𝑘𝑙}}=$$\sum {𝐿}_{{𝑐}_{𝑖𝑗𝑘𝑙}{𝑈}_{𝑖𝑗},{𝑈}_{𝑘𝑙}}$, a sum of maps of the form ${𝐿}_{𝐴,𝐵}$.

The transpose sends ${𝑈}_{𝑗𝑘}$ to ${𝑈}_{𝑘𝑗}$, so it is the sum of the maps with $𝑖=𝑘$ and $𝑙=𝑗$:

Directly, ${𝑈}_{𝑘𝑗}𝑋{𝑈}_{𝑘𝑗}=$${𝑥}_{𝑗𝑘}{𝑈}_{𝑘𝑗}$, and summing gives the matrix with $(𝑘,𝑗)$ entry ${𝑥}_{𝑗𝑘}$. As a check against (b), $tr𝜏=$${\sum }_{𝑗,𝑘}{(tr{𝑈}_{𝑘𝑗})}^2=$${\sum }_{𝑘}1=$$𝑛\ast 1$. So the transpose is a sum of ${𝑛}^2$ maps $𝑋\mapsto 𝐴𝑋𝐵$, although for $𝑛\ge 2$ it is not one of them: if ${𝑋}^{\text{𝖳}}=𝐴𝑋𝐵$ for all $𝑋$, then ${𝑈}_{11}=$$(𝐴{𝐞}_1)({𝐞}_1^{\text{𝖳}}𝐵)$ and ${𝑈}_{21}=$$(𝐴{𝐞}_1)({𝐞}_2^{\text{𝖳}}𝐵)$, by taking $𝑋={𝑈}_{11}$ and $𝑋={𝑈}_{12}$. Comparing first columns, $𝐴{𝐞}_1\ne \mathrm{𝟎}$ would be a multiple of both ${𝐞}_1$ and ${𝐞}_2$, which is impossible since they are independent.

Throughout, by induction on $𝑘$, ${𝐷}^{𝑘}({𝑥}^{𝑚})=$$(𝑚(𝑚-1)\cdots (𝑚-𝑘+1)\ast 1){𝑥}^{𝑚-𝑘}$ for $𝑘\le 𝑚$, and ${𝐷}^{𝑘}({𝑥}^{𝑚})=\mathrm{𝟎}$ for $𝑘>𝑚$ (Example 2.1.9).

(a) ${𝑆}_{𝑐}$ maps ${\mathbb{Q}[𝑥]}_{\le 𝑛}$ into itself, since ${(𝑥+𝑐)}^{𝑚}$ has degree $𝑚$, and so does every ${𝐷}^{𝑘}$. Both sides of the claim are linear (Proposition 2.1.23, Proposition 2.1.17), so by Corollary 2.2.6 it suffices to compare them on ${𝑥}^{𝑚}$ for $0\le 𝑚\le 𝑛$. For $𝑘\le 𝑚$, $𝑚(𝑚-1)\cdots (𝑚-𝑘+1)/𝑘!=$$\left(\genfrac{}{}{0ex}{}{𝑚}{𝑘}\right)$, and the terms with $𝑘>𝑚$ vanish, so

(b) ${𝑆}_1{𝑆}_{-1}=$${𝑆}_{-1}{𝑆}_1=$${𝑆}_0=$$\text{id}$, and both maps preserve ${\mathbb{Q}[𝑥]}_{\le 𝑛}$. By Technique 2.5.16, ${𝑆}_1$ is invertible on ${\mathbb{Q}[𝑥]}_{\le 𝑛}$ with inverse ${𝑆}_{-1}$, and by (a) with $𝑐=-1$,

(c) The case $𝑛<𝑝$. For $𝑘\le 𝑛<𝑝$, the scalar $𝑘!\ast 1$ is a product of the nonzero elements $1,2,\dots ,𝑘$ of ${𝔽}_{𝑝}$, so it is invertible. The integer identity $𝑚(𝑚-1)\cdots (𝑚-𝑘+1)=$$𝑘!\left(\genfrac{}{}{0ex}{}{𝑚}{𝑘}\right)$ holds in ${𝔽}_{𝑝}$ after reduction, so ${(𝑘!)}^{-1}{𝐷}^{𝑘}({𝑥}^{𝑚})=$$\left(\genfrac{}{}{0ex}{}{𝑚}{𝑘}\right){𝑥}^{𝑚-𝑘}$, and the computation of (a) with $𝑐=1$ goes through unchanged.

The case $𝑛\ge 𝑝$. Suppose ${𝑆}_1=$${\sum }_{𝑘=0}^{𝑟}{𝑎}_{𝑘}{𝐷}^{𝑘}$ on ${𝔽}_{𝑝}{[𝑥]}_{\le 𝑛}$. The polynomial ${𝑥}^{𝑝}$ lies in this space. Since $𝐷({𝑥}^{𝑝})=$$(𝑝\ast 1){𝑥}^{𝑝-1}=$$\mathrm{𝟎}$, every ${𝐷}^{𝑘}({𝑥}^{𝑝})$ with $𝑘\ge 1$ is $\mathrm{𝟎}$, and the right side sends ${𝑥}^{𝑝}$ to ${𝑎}_0{𝑥}^{𝑝}$. On the other hand, the coefficients $\left(\genfrac{}{}{0ex}{}{𝑝}{𝑘}\right)$ with $0<𝑘<𝑝$ are $0$ in ${𝔽}_{𝑝}$, so

So ${𝑥}^{𝑝}+1={𝑎}_0{𝑥}^{𝑝}$, and the constant terms give $1=0$, a contradiction.

Over ${𝔽}_{𝑝}$ the derivative cannot see ${𝑥}^{𝑝}$, but the shift moves it. So Taylor’s formula is a statement about fields in which the factorials can be inverted, and it fails exactly when the space contains ${𝑥}^{𝑝}$.

(a) If $𝑇(𝑓)=𝑞𝑓$ for all $𝑓$, then $𝑇(1)=𝑞$ and $𝑇(𝑀𝑓)=$$𝑞𝑥𝑓=$$𝑥𝑞𝑓=$$𝑀(𝑇(𝑓))$ by commutativity. Conversely, suppose $𝑇𝑀=𝑀𝑇$. By induction, $𝑇{𝑀}^{𝑘}={𝑀}^{𝑘}𝑇$ for every $𝑘\ge 0$: if $𝑇{𝑀}^{𝑘}={𝑀}^{𝑘}𝑇$, then $𝑇{𝑀}^{𝑘+1}=$$(𝑇{𝑀}^{𝑘})𝑀=$${𝑀}^{𝑘}𝑇𝑀=$${𝑀}^{𝑘}𝑀𝑇$. With $𝑞=𝑇(1)$,

Multiplication by $𝑞$ is linear, by distributivity and $𝑐(𝑞𝑓)=𝑞(𝑐𝑓)$, and it agrees with $𝑇$ on every monomial. By the uniqueness in Proposition 2.2.20, $𝑇(𝑓)=𝑞𝑓$ for all $𝑓$.

(b) If $𝑞=\mathrm{𝟎}$, then $𝑇=0$, which is not injective. If $𝑞\ne \mathrm{𝟎}$ and $𝑓\ne \mathrm{𝟎}$, then $deg(𝑞𝑓)=$$deg𝑞+$$deg𝑓\ge$$0$, so $𝑞𝑓\ne \mathrm{𝟎}$; by Theorem 2.6.11, $𝑇$ is injective.

If $𝑞=𝑐$ is a nonzero constant, then $𝑇=𝑐\text{id}$ is surjective. Conversely, if $𝑇$ is surjective, then $1=𝑞𝑔$ for some $𝑔$. Both $𝑞$ and $𝑔$ are nonzero, so $deg𝑞+$$deg𝑔=$$deg1=$$0$, which forces $deg𝑞=0$: $𝑞$ is a nonzero constant. For instance, multiplication by $𝑥$ is injective but not surjective, as in Example 2.6.4(e).

(c) By (a), $𝑇$ commutes with $𝑀$ exactly when it is multiplication by some $𝑞$. For such $𝑇$, the product rule gives $𝐷(𝑇(𝑓))=$$𝐷(𝑞𝑓)=$$𝐷(𝑞)𝑓+$$𝑞𝐷(𝑓)=$$𝐷(𝑞)𝑓+$$𝑇(𝐷(𝑓))$. So $𝐷𝑇=𝑇𝐷$ if and only if $𝐷(𝑞)𝑓=\mathrm{𝟎}$ for every $𝑓$. Taking $𝑓=1$ shows that this forces $𝐷(𝑞)=\mathrm{𝟎}$, and conversely $𝐷(𝑞)=\mathrm{𝟎}$ makes it hold. So the operators commuting with $𝑀$ and $𝐷$ are the multiplications by the polynomials in $ker𝐷$, which Example 2.6.27 computed.

• If $𝐹$ has characteristic $0$, $ker𝐷$ consists of the constants, so $𝑇=𝑐\text{id}$ for some $𝑐\in 𝐹$.
• If $𝐹={𝔽}_{𝑝}$, $ker𝐷$ consists of the polynomials ${𝑏}_0+$${𝑏}_1{𝑥}^{𝑝}+$${𝑏}_2{𝑥}^{2𝑝}+$$\cdots$, so $𝑇$ is multiplication by a polynomial in ${𝑥}^{𝑝}$. For instance $𝑓\mapsto {𝑥}^{𝑝}𝑓$ commutes with $𝐷$, because $𝐷({𝑥}^{𝑝})=$$(𝑝\ast 1){𝑥}^{𝑝-1}=$$\mathrm{𝟎}$.

(a) Basis. Let $𝑓\in 𝑉$. By Proposition 3.4.27 with ${𝑑}_{𝑖}=𝑓({𝑐}_{𝑖})$, the polynomial ${\sum }_{𝑖}𝑓({𝑐}_{𝑖}){𝐿}_{𝑖}$ is the unique element of $𝑉$ with the values $𝑓({𝑐}_0),$$\dots ,$$𝑓({𝑐}_{𝑛})$, and $𝑓$ is such an element, so $𝑓=$${\sum }_{𝑖}𝑓({𝑐}_{𝑖}){𝐿}_{𝑖}$. Hence the ${𝐿}_{𝑖}$ span $𝑉$. If ${\sum }_{𝑖}{𝑎}_{𝑖}{𝐿}_{𝑖}=\mathrm{𝟎}$, then evaluating at ${𝑐}_{𝑘}$, which is linear (Proposition 1.2.18(c)), gives ${𝑎}_{𝑘}=0$. So ${𝐿}_0,\dots ,{𝐿}_{𝑛}$ is a basis of $𝑉$, and the coordinates of $𝑓$ in it are $(𝑓({𝑐}_0),\dots ,𝑓({𝑐}_{𝑛}))$.

The formula. Each ${ev}_{{𝑐}_{𝑖}}$ is linear (Example 2.1.9), so $𝜓=$${\sum }_{𝑖}𝜑({𝐿}_{𝑖}){ev}_{{𝑐}_{𝑖}}$ is linear (Proposition 2.1.23). On a basis vector,

since only the term $𝑖=𝑘$ survives. By Corollary 2.2.6, $𝜓=𝜑$. So the evaluations span $\mathrm{L}(𝑉,𝐹)$. If ${\sum }_{𝑖}{𝑎}_{𝑖}{ev}_{{𝑐}_{𝑖}}$ is the zero map, applying it to ${𝐿}_{𝑘}$ gives ${𝑎}_{𝑘}=0$. So ${ev}_{{𝑐}_0},\dots ,{ev}_{{𝑐}_{𝑛}}$ is a basis of $\mathrm{L}(𝑉,𝐹)$, and the coefficients of $𝜑$ in it are the values $𝜑({𝐿}_{𝑖})$.

(b) Since $2\ne 0$, the points $0,1,2$ are distinct. The polynomials

have the required values: ${𝐿}_0$ takes the values $1,$$1-$$\frac{3}{2}+$$\frac{1}{2}=$$0,$$1-$$3+$$2=$$0$ at $0,1,2$; ${𝐿}_1$ takes $0,1,4-4=0$; and ${𝐿}_2$ takes $0,0,-1+2=1$. By the uniqueness in Definition 3.4.26, these are the Lagrange polynomials. The map $𝜑(𝑓)=$${𝑓}^{\prime }(0)=$${ev}_0(𝐷(𝑓))$ is linear, and for $𝑓=𝑎+𝑏𝑥+𝑐{𝑥}^2$, $𝐷(𝑓)=𝑏+2𝑐𝑥$, so ${𝑓}^{\prime }(0)=𝑏$ is the coefficient of $𝑥$. By (a), the formula holds with ${𝑤}_{𝑖}=𝜑({𝐿}_{𝑖})$, and with no other scalars, since the evaluations are independent:

Over ${𝔽}_3$, $2^{-1}=2$, so $-\frac{3}{2}=0$ and $-\frac{1}{2}=-2=1$, and the formula reads ${𝑓}^{\prime }(0)=$$2𝑓(1)+$$𝑓(2)$. As a check over ${𝔽}_3$: for $𝑓=1$, $𝑥$, ${𝑥}^2$ the right side is $2+1=0$, $2+2=1$, $2+4=0$, which are the coefficients of $𝑥$.

(c) Let $𝑓={𝑥}^3-3{𝑥}^2+2𝑥$. Then $𝑓(0)=0$, $𝑓(1)=1-3+2=0$, $𝑓(2)=8-12+4=0$, and ${𝑓}^{\prime }(0)=2$, the coefficient of $𝑥$. Any formula ${𝑓}^{\prime }(0)=$${𝑤}_0𝑓(0)+$${𝑤}_1𝑓(1)+$${𝑤}_2𝑓(2)$ would give $2=0$. The three values cannot see a cubic that vanishes at all three points; on ${\mathbb{Q}[𝑥]}_{\le 3}$ the evaluations at three points span only a part of $\mathrm{L}({\mathbb{Q}[𝑥]}_{\le 3},\mathbb{Q})$.

Write $𝑡=𝑎+𝑑$ and $\mathrm{\Delta }=𝑎𝑑-𝑏𝑐$, so that ${𝐴}^2=𝑡𝐴-\mathrm{\Delta }{𝐼}_2$.

(a) Suppose ${𝐴}^2=𝐴$. Then $𝐴=𝑡𝐴-\mathrm{\Delta }{𝐼}_2$, that is, $(𝑡-1)𝐴=\mathrm{\Delta }{𝐼}_2$. If $𝑡\ne 1$, then $𝐴=$${(𝑡-1)}^{-1}\mathrm{\Delta }{𝐼}_2=$$𝜆{𝐼}_2$ for a scalar $𝜆$, and ${𝐴}^2=𝐴$ gives ${𝜆}^2=𝜆$, so $𝜆\in \{0,1\}$ and $𝐴=0$ or $𝐴={𝐼}_2$. If $𝑡=1$, then $\mathrm{\Delta }{𝐼}_2=0$, so $\mathrm{\Delta }=0$. Conversely, $0$ and ${𝐼}_2$ are idempotent, and if $𝑡=1$ and $\mathrm{\Delta }=0$, then ${𝐴}^2=𝐴-0=𝐴$. (The three cases do not overlap: $tr0=0$ and $tr{𝐼}_2=1+1$, and neither is $1$.)