I'm trying to prove/disprove a homework problem that is the title question. I'm not looking for an explicit answer, just some direction. So, I've been reading Axler's book Linear Algebra Done Right and from what I understand,
every operator on $V$ has a diagonal matrix with respect to some orthonormal bases of $V$, provided that we are permitted to use two different bases rather than a single basis as customary when working with operators.
So that if we let $( e_1, ..., e_n )$ and $( f_1, ..., f_n )$ be orthonormal bases of $V$ then the matrix of $T$, an operator on $V$, is
$$M(T,(e_1,...,e_n),(f_1,...,f_n)) = \begin{bmatrix} d_{1} & 0 & \cdots & 0 \\ 0 & d_{2} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & d_{n} \end{bmatrix}$$
So, my thinking is that the matrix of $T^2$ is just the matrix above composed with itself and so would be
$\begin{bmatrix} d^2_{1} & 0 & \cdots & 0 \\ 0 & d^2_{2} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & d^2_{n} \end{bmatrix}$
Am I on the right track or am I missing something?