I have been reviewing my Linear Algebra using Friedberg, Insel, and Spences' Linear Algebra 4th Edition and I found something curious in the exercises. In Section 2.2, exercise 16 states:
Let $V$ and $W$ be vector spaces such that $\dim(V) = \dim(W)$, and let $T: V \rightarrow W$ be linear. Show that there exist ordered bases $\beta$ and $\gamma$ for $V$ and $W$, respectively, such that $[T]_\beta^\gamma$ is a diagonal matrix.
My question is this: If I let $V = W$, then does this exercise say that every linear operator on a finite-dimensional vector space is diagonalizable? I know that this is not true, but I can't see where the issue is with the exercise statement. My gut feeling is that $\beta$ and $\gamma$ will always be distinct if $T$ is not diagonalizable, but I'm not sure if that's true.