Exercise 5.4.24

Let $T$ be an operator on $V$ and $W$ be a nontrivial $T$-invariant subspace. Also let $E_{\lambda }$ be the eigenspace of $V$ corresponding to the eigenvalue $\lambda$. We set $W_{\lambda }=E_{\lambda }\cap W$ to be the eigenspace of $T_W$ corresponding to the eigenvalue $\lambda$. We may find a basis ${\beta }_{\lambda }$ for $W_{\lambda }$ and try to show that $\beta ={\text{∪}}_{\lambda }{\beta }_{\lambda }$ is a basis for $W$. The set $\beta$ is linearly independent by Theorem 5.8. Since $T$ is diagonalizable, every vector could be written as a linear combination of eigenvectors corresponding to distinct eigenvalues, so are those vectors in $W$. But by the previous exercise, we know that those eigenvectors used to give a linear combination to elements in $W$ must also be in $W$. This means every element in $W$ is a linear combination of $\beta$. So $\beta$ is a basis for $W$ consisting of eigenvectors.