# Linear transformations with equal matrices on different bases

Let V be a finite dimensional vector space over $\mathbb{F}$ , and let $T,S:V\to V$ be linear transformations and $B,C$ ordered bases for $V$. Assume that $$\left[T\right]_{B}^{B}=\left[S^{2}\right]_{C}^{C}$$ (Where $[T]^B_B$ is the matrix of the linear transformation $T$ with respect to bases $B$ and $B$. Prove or disprove: There is a linear transformation $R:V\to V$ such that $T=R^{2}$.

My feeling is that it's correct though I haven't been able to prove it, I've tried switching the bases by defining

$$R\left(v\right)=\left[Id_{v}\right]_{B}^{C}\left[S\right]_{C}^{C}\left[v\right]_{C}$$

(Where $Id_V$ is the identity of $V$ and $[v]_C$ is the coordinates of $v$ with respect to $C$)

But then I am stuck at proving equality as the bases stay different, i.e.

$$T\left(v\right)=\left[T\right]_{B}^{B}\left[v\right]_{B}=\left[S^2\right]^C_C[v]_B$$ and I'm not sure how to switch the base.

Any ideas?

Well $[T]_B=[S^2]_C=([S]_C)^2$ so consider the map $R$ for which $[R]_B=[S]_C$.
Every matrix defines a linear map wrt a chosen basis, so in particular there exists a linear map $R$ which acts by the matrix $[S]_C$ wrt the basis $B$. And two linear maps that act by the same matrix wrt a fixed basis are the same map - thus $[T]_B=[R^2]_B$ implies $T=R^2$.