Exercise 5.4.29

We use the notation given in Hint of the previous exercise again. By Exercise 5.4.24 we may find a basis $\gamma$ for $W$ such that ${[T_W]}_{\gamma }$ is diagonal. For each eigenvalue $\lambda$, we may pick the corresponding eigenvectors in $\gamma$ and extend it to be a basis for the corresponding eigenspace. By collecting all these bases, we may form a basis $\beta$ for $V$ who is the union of these bases. So we know that ${[T]}_{\beta }$ is diagonal. Hence the matrix $B_3$ is also diagonal. Since we’ve proven that $B_3={[\overline{T}]}_{\alpha }$, we know that $\overline{T}$ is diagonalizable.