Defining $\mathcal{A}(S) = TS$ implies $\mathcal{A}$ and $T$ share the same eigenvalues. Need help with proof step.

The following exercise comes from Linear Algebra Done Right, 4th Edition, Sheldon Axler in Section 5A, exercise number 37.

Suppose $$V$$ is finite-dimensional and $$T \in \mathcal{L}(V)$$. Define $$\mathcal{A} \in \mathcal{L}(\mathcal{L}(V))$$ by

$$\mathcal{A}(S) = TS$$

for each $$S \in \mathcal{L}(V)$$. Prove that the set of eigenvalues of $$T$$ equals the set of eigenvalues of $$\mathcal{A}$$.

Here, $$\mathcal{L}(V)$$ is the set (vector space) of linear operators defined on a vector space $$V$$.

This is how I thought of the problem:

Let $$E_T$$ be the set of eigenvalues of $$T$$ and let $$E_{\mathcal{A}}$$ be the set of eigenvalues of $$\mathcal{A}$$. We want to show $$E_T = E_{\mathcal{A}}$$, which means showing

$$\lambda \in E_T$$ such that there exists a nonzero $$v \in V$$ where $$Tv = \lambda v$$ if and only if $$\lambda \in E_{\mathcal{A}}$$ such that there exists a nonzero $$S \in \mathcal{L}(V)$$ where $$\mathcal{A}(S) = \lambda S$$.

Since $$V$$ is finite-dimensional, it has a basis $$v_1, \ldots, v_n$$.

Here's what I have so far showing the backward direction:

Let $$\lambda \in E_{\mathcal{A}}$$. This implies there exists a nonzero $$S \in \mathcal{L}(V)$$ such that $$\mathcal{A}(S) = \lambda S$$. But by definition of $$\mathcal{A}$$, this implies $$\lambda S = TS$$. That is, for all $$v \in V$$, we have

$$\lambda (Sv) = T(Sv)$$

I have to show $$Sv$$ is nonzero, but this is where I'm a bit confused. Do I have to show $$Sv \neq 0$$ for every nonzero $$v$$? Or can I get away with choosing some nonzero vector (say $$v_1$$ since it's in the basis) and conclude that there exists a nonzero $$Sv_1 \in V$$ such that $$\lambda (Sv_1) = T(Sv_1)$$?

• You choose $\lambda \in E_{\mathcal A}$. You want to prove that $\lambda \in E_{T}$. By definition, you want to exhibit a non zero vector $w \in V$ such that $Tw = \lambda w$. So far, you have proved that for all $v\in V$, you have $TSv = \lambda Sv$. Thus, it is very tempting to choose $w$ of the form $Sv$ for some $v \in V$. The problem is, can you choose a $v$ such that $w := Sv$ is non zero? If so, for what reason? If you can answer this, you are done with the backward direction Commented May 21 at 2:57
• @Suzet Thank you for commenting. Well, we have that $S \neq 0$. So, there must exist some $v$ such that $Sv \neq 0$ since otherwise, $S$ would be the zero operator, which is a contradiction. Is this reasoning correct? Commented May 21 at 3:14
• Exactly, this is the expected justification :) Commented May 21 at 3:16
• @Suzet Ah, of course! Thank you for your help. Commented May 21 at 3:19

As discussed in the comments, the backward direction has been shown. Now I show the forward direction.

Let $$\lambda \in E_T$$. This implies there exists a nonzero $$u \in V$$ such that

$$Tu = \lambda u$$

We want to show there exists a nonzero $$S \in \mathcal{L}(V)$$ such that $$\mathcal{A}(S) = \lambda S$$. Since $$\mathcal{A}(S) = TS$$, this is equivalent to showing $$(TS)v = (\lambda S)v$$ for all $$v \in V$$.

With $$v_1, \ldots, v_n$$ a basis of $$V$$, we use the linear map lemma to define $$S \in \mathcal{L}(V)$$ by

$$Sv_j = u$$

for all $$j = 1, \ldots, n$$. Showing $$S$$ is in fact linear is easy to show, so I leave it out. This definition of $$S$$ implies that it is nonzero (since otherwise, $$S = 0$$ implies $$Sv_j = 0v_j = 0 = u$$, contradicting that $$u$$ is nonzero).

Moreover, we have that for any $$v \in V$$,

$$v = a_1v_1 + \cdots + a_nv_n$$

for some scalars $$a_1, \ldots, a_n \in \mathbb{F}$$ (here, $$\mathbb{F}$$ is the set of real or complex numbers). Now, observe that

\begin{align} (TS)v = T(Sv) = T(S(a_1v_1 + \cdots +a_nv_n)) &= T(a_1Sv_1+ \cdots + a_nSv_n) \\ &=T(a_1u + \cdots + a_nu) \\ &= a_1Tu + \cdots + a_nTu \\ &= a_1 \lambda u + \cdots + a_n \lambda u \\ &= a_1 \lambda Sv_1 + \cdots + a_n \lambda Sv_n \\ &= \lambda S (a_1v_1 + \cdots + a_nv_n) \\ &= \lambda (Sv) \\ &= (\lambda S)v \end{align}

The above shows that $$(TS)v = (\lambda S)v$$ for any $$v \in V$$, that is, $$\mathcal{A}(S) = \lambda S$$ where $$S$$ is nonzero. This implies $$\lambda$$ is an eigenvalue of $$\mathcal{A}$$ and hence, $$\lambda \in E_\mathcal{A}$$.

By showing both directions, we have shown $$E_T = E_\mathcal{A}$$.