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

Question

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 $\mathcal{A}$.

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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 $\lambda (Sv_1) = T(Sv_1)$?

Answer 1

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 $E_T = E_\mathcal{A}$.