Does $V\rightarrow W$ isomorphism map a $T:V\rightarrow V$ eigenvector to an $S:W\rightarrow W$ eigenvector?

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How would go about proving the following lemma?

Lemma: Let $V,W$ be vector spaces over $\mathbb{F}$, $\alpha:V \rightarrow W$ be an isomorphism, and $T:V \rightarrow V, S:W \rightarrow W$ be $\mathbb{F}$ linear maps. Then $v_1,...,v_n $ is an $\mathbb{F}$ basis of $V$ consisting of eigenvectors of $T$ $\iff$ $\alpha(v_1),...,\alpha(v_n) $ is an $\mathbb{F}$ basis of $W$ consisting of eigenvectors of $S$.

It is clear to me that $\alpha(v_1),...,\alpha(v_n)$ is a basis on $W$. However, I'm stuck on trying to show they are also eigenvectors.

I am attempting to use this as a lemma to prove a theorem covered in my lin algebra class:

Let $V$ be a vector space over $\mathbb{C}$ and $T:V \rightarrow V $ be $ \mathbb{C}$-linear. Then $T:V \rightarrow V$ is diagonalizable $\iff \forall \lambda \in \sigma(T)$, ${\rm a.mult}(\lambda) = {\rm g.mult}(\lambda)$.

In the proof given by my professor, he assumes that $V = \mathbb{C}^n$.

Edit: This lemma is obviously wrong as helpfully pointed out by @Drew Brady.

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If you add the additional condition that $S = \alpha \circ T \circ \alpha^{-1}$, the result holds immediately, since $$ S (\alpha v) = \alpha \circ Tv = \alpha( \lambda v) = \lambda \alpha(v), $$ for any eigenvector-value pair $(\lambda, v)$ of $T$.