Verification of proof of one direction of the complex spectral theorem.

Question

In trying to prove the complex spectral theorem I came up with a proof that is far different from the one in my book.

$\mathbf{Theorem :}$ Suppose that $V$ is a complex inner product space and $T \in L(V)$. Then $V$ has an orthogonal basis consisting of eigenvectors of $T$ if $T$ is normal.

$\mathbf{Proof: }$ Assume that $T$ is normal. Since $V$ is complex, $T$ has an eigenvalue $\lambda$ with an associated eigenvector $u_1$. Extend $u_1$ to an orthogonal basis of $V$ (Gran-Schmidt), $B_v = (u_1, v_1, ... , v_{n-1})$. Now notice that for any $i = 1,2,...,n-1$ we have

$$ = $<\lambda u_1, v_i>$ = $$

Where the second equality comes from the properties of an inner product.

But also notice that $$ = $$, where $T^*$ is the adjoint of $T$. So $T^*v_i = \overline{\lambda}v_i$. So $v_i$ is a eigenvector of $T^*$. By a lemma (proved in my book), $v_i$ is an eigenvector of $T$. So $B_v$ is a basis of orthogonal eigenvectors.

Does this seem correct? Thank you.

Answer 1

This proof cannot be correct, for looking at the matrix $$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix} $$ we could pick $u_1 = e_1$, but $v_1 = s e_2 + s e_3$ and $v_2 = s e_2 - s e_3$, where $ s= \sqrt{\frac{1}{2}}$.

Your proof would then show that $v_1$ and $v_2$ are eigenvectors, which they are not.