Exercise 7.4.5

If the rational canonical form of $T$ is a diagonal matrix, then $T$ is diagonalizable naturally. Conversely, if $T$ is diagonalizable, then the characteristic polynomial of $T$ splits and $E_{\lambda }=K_{{\varphi }_{\lambda }}$, where ${\varphi }_{\lambda }=t-\lambda$, for each eigenvalue $\lambda$. This means each cyclic basis in $K_{{\varphi }_{\lambda }}$ is of size $1$. That is, a rational canonical basis consisting of eigenvectors. So the rational canonical form of $T$ is a diagonal matrix.