Exercise 6.4.8

By Theorem 6.16 we know that $T$ is diagonalizable. Also, by Exercise 5.4.24 we know that $T_W$ is also diagonalizable. This means that there’s a basis for $W$ consisting of eigenvectors of $T$. If $x$ is a eigenvectors of $T$, then $x$ is also a eigenvector of $T^{\text{∗}}$ since $T$ is normal. This means that there’s a basis for $W$ consisting of eigenvectors of $T^{\text{∗}}$. So $W$ is also $T$-invariant.

Here is a better solution. Take B to be a basis for W with $T(B)=\lambda$ B. Since T is normal, we have:

Thus, we see that $T^{\ast }x$ is still within W.