Exercise 8.B.10

Proof. Let $v_1,\dots ,v_n$ denote a basis of $V$ consisting of generalized eigenvectors of $T$ (which exists by 8.23). Define $⟨\cdot ,\cdot ⟩:V\times V\to ℂ$ by

where the $a$’s and $b$’s are complex numbers. You can check that $\text{⟨⋅,⋅⟩}$ is a well-defined inner product on $V$. Thus $v_1,\dots ,v_n$ is an orthonormal basis of $V$. Moreover, the generalized eigenspaces of $T$ are orthogonal to each other. This implies that, if $v\in G(\beta ,T)$, then

where $P_{G(\alpha ,T)}$ is the orthogonal projection of $V$ onto $G(\alpha ,T)$.

Let ${\lambda }_1,\dots ,{\lambda }_m$ denote the distinct eigenvalues of $T$. We have

For each $j=1,\dots ,m$, we can write $T{\text{|}}_{G({\lambda }_j,T)}={\lambda }_{j}I+N_j$ where $N_j$ is a nilpotent operator under which $G({\lambda }_j,T)$ is invariant (see 8.21 (c)). Therefore,

Fix $k\in \{1,\dots ,n\}$. Then $v_k\in G({\lambda }_j,T)$ for some $j\in \{1,\mathit{\dots m}\}$. $\text{(∗)}$ shows that $(4)$ maps $v_k$ to ${\lambda }_j{v}_k$. Hence $v_1,\dots ,v_n$ is a basis of eigenvectors of $(4)$ and so $(4)$ is diagonalizable. $\text{(∗)}$ also shows that $(5)$ maps $v_k$ to $N_j{v}_k$. But $G({\lambda }_j,T)$ is invariant under $N_j$, so $\text{(∗)}$ actually implies that $(5)$ raised to the power of $\dim <!--\; FUNCTION\; APPLICATION\; -->V$ maps $v_k$ to $N_j^{\dim <!--\; FUNCTION\; APPLICATION\; -->V}{v}_k$ which equals $0$. Therefore $(5)$ is nilpotent. It is easy to see that $(4)$ and $(5)$ commute (they map $v_k$ to ${\lambda }_j{N}_j{v}_k$, no matter the order), which completes the proof. □