# Exercise 5.B.19 from "Linear Algebra Done Right", Sheldon Axler, 4th edition.

The following is exercise 5.B.19 from Linear Algebra Done Right, Sheldon Axler, fourth edition.

Suppose $$V$$ is finite-dimensional and $$T \in \mathcal{L}(V)$$. Let $$\mathcal{E}$$ be the subspace of $$\mathcal{L}(V)$$ defined by

$$\mathcal{E} = \{q(T) : q \in \mathcal{P}(\mathbf{F})\}$$

Prove that $$\dim \mathcal{E}$$ equals the degree of the minimal polynomial of $$T$$.

I'm having quite some trouble with this exercise. Letting $$p \in \mathcal{P}(\mathbf{F})$$ be the minimal polynomial of $$T$$, here are some facts I know:

(1) $$p(T) = 0$$, $$p(T) \in \mathcal{E}$$, and $$\deg p \leq \dim V$$.

(2) $$\mathcal{E} \neq \mathcal{L}(V)$$, implying $$\mathcal{L}(V) = \mathcal{E} \oplus W$$ for some subspace $$W$$ of $$\mathcal{L}(V)$$. This implies

\begin{align} \dim \mathcal{E} &= \dim \mathcal{L}(V) - \dim W \\ &= (\dim V)^2 - \dim W \end{align}

And some strategies I tried:

(1) Picked an arbitrary basis of $$\mathcal{E}$$, say $$q_1(T), \ldots, q_k(T)$$. Then

$$p(T) = \alpha_1 q_1(T) + \cdots + \alpha_k q_k(T) = (\alpha_1 q_1 + \cdots + \alpha_k q_k)(T) = 0$$

This implies there is some $$s \in \mathcal{P}(\mathbf{F})$$ such that $$p = (\alpha_1 q_1 + \cdots + \alpha_k q_k)s$$. Can I imply anything useful from here? I assume not because we must have that $$\alpha_1 = \cdots = \alpha_k = 0$$.

(2) Tried showing that $$\dim \mathcal{E} \leq \deg p$$ and $$\dim \mathcal{E} \geq \deg p$$, but found this hard to do.

(3) Thinking about the null space and range of $$p$$ or any $$q$$, but I'm not sure how I should be thinking about them.

Could someone lend me a helping hand? Will any of the above facts and strategies help me?

• Show that $I, T, T^2, \ldots, T^{n-1}, n$ being the degree of the minimum polynomial, forms a basis. Independence is clear. For span remainder theorem is useful. Commented Jul 18 at 0:39

Consider the following three results found in the book:

3.107 $$\$$ null space and range of $$\widetilde{T}$$

Suppose $$T \in \mathcal{L}(V,W)$$. Then [...]

(d) $$V/(\operatorname{null} T)$$ and $$\operatorname{range} T$$ are isomorphic vector spaces.

Exercises 4

13 $$\$$ Suppose $$p \in \mathcal{P}(\mathbf{F})$$ with $$p \neq 0$$. Let $$U = \{pq : q \in \mathcal{P}(\mathbf{F})\}$$.

(a) Show that $$\dim \mathcal{P}(\mathbf{F})/U = \deg p$$. [...]

5.29 $$\$$ $$q(T)=0 \iff q$$ is a polynomial multiple of the minimal polynomial

Suppose $$V$$ is finite dimensional, $$T \in \mathcal{L}(V)$$, and $$q \in \mathcal{P}(\mathbf{F})$$. Then $$q(T)=0$$ if and only if $$q$$ is a polynomial multiple of the minimal polynomial of $$T$$.

Now, for your problem, consider the linear map $$E \colon \mathcal{P}(\mathbf{F}) \to \mathcal{L}(V)$$ defined by $$E(q) = q(T).$$ By 5.29, $$\operatorname{null} E = \{pq : q \in \mathcal{P}(\mathbf{F})\}$$ where $$p$$ is the minimal polynomial of $$T$$. Also, it is easy to see that $$\operatorname{range} E = \mathcal{E}$$.

Hence, by 3.107, $$\mathcal{E}$$ is isomorphic to $$\mathcal{P}(\mathbf{F})/\{pq : q \in \mathcal{P}(\mathbf{F})\}$$; so, by exercise 4.13, $$\dim \mathcal{E} = \deg p.$$

• Perfect, thank you! Wish I had looked at chapter 4 exercises more carefully. Commented Jul 18 at 2:03

When are two polynomials equivalent as functions of $$T$$? Exactly when they are congruent modulo $$p$$, the minimal polynomial of $$T$$ (ie when their difference is a polynomial multiple of $$p$$) This is not so hard to show, and it is an important fact.

Can you show that this quotient space of polynomials is of the right dimension? Just realize that, since every polynomial is congruent to its remainder on division by $$p$$, there is a system of class representatives with no polynomials of degree greater than the degree of $$p$$.

Finally, since $$p$$ is the minimal polynomial, all polynomials of degree less than $$p$$ are not congruent. This means that our quotient space is really just (obviously isomorphic to) the space of polynomials of degree less than the degree of $$p$$, which has the appropriate dimension.

Finally, we can construct the obvious isomorphism between this space of polynomials and our subspace.