Making matrices look the same w.r.t different basis's

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Suppose that $T, S \in L(V )$.

Find necessary and sufficient conditions for there to exist bases $\alpha , \alpha ' , \beta, \beta ' $ such that $ [T]_{\alpha}^{\beta} =[S]_{\alpha '}^{\beta '}$

i would like to make one of these matrices a diagonal matrices w.r.t the new basis's which i know how to do but im not sure how to also adjust the second one so it is exactly the same with the new basis...

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It is necessary and sufficient that $S$ and $T$ have the same rank.

If $T$ has rank $k$, you may find bases $\alpha,\beta$ such that $[T]_\alpha^\beta$ has the canonical form $[T]_\alpha^\beta = I_k \oplus 0$. Indeed, take a basis $(e_i)$ for $\operatorname{im}(T)$ and complete it to a basis $\beta$ for $V$. Then pick $k$ linearly independent preimage vectors $(T^{-1}(e_i))$ and complete them to a basis $\alpha$ for $V$ using basis vectors for $\ker(T)$. Hence, if $S$ and $T$ have the same rank, you may find pairs of bases for both such that both have the same canonical form.

The converse follows from the well-known result that the rank of a linear map coincides with the rank of any of its matrix representations.