Let $V$ be a hermitian space and $f:V\to V$ is an operator. Show that $f$ is normal operator iff any eigenvector of $f$ is also eigenvector of $f^*$.
My approach:
$\Rightarrow$ Suppose that $f$ is normal operator then one can show that for any scalar $\mu$ the operator $f-\mu\cdot \text{id}$ is also normal. Let $x$ is the eigenvector of $f$ with eigenvalue $\lambda$, i.e. $f(x)=\lambda x$. Then:
$$0=((f-\lambda \cdot \text{id})^*(f-\lambda \cdot \text{id})(x),x)=((f^*-\overline{\lambda} \cdot \text{id})(f-\lambda \cdot \text{id})(x),x)=((f-\lambda \cdot \text{id})(f^*-\overline{\lambda} \cdot \text{id})(x),x)=((f^*-\overline{\lambda} \cdot \text{id})(x),(f^*-\overline{\lambda} \cdot \text{id})(x))$$ which means that $(f^*-\overline{\lambda} \cdot \text{id})(x)=0$, i.e. $f^*(x)=\overline{\lambda}x$. So we have shown that $x$ is the eigenvector for $f^*$ with eigenvalue $\overline{\lambda}$.
$\Leftarrow$ I was not able to prove this direction but anyway let me show what I have done so far. Let $\chi_f(t)\in \mathbb{C}[t]$ is characteristic polynomial of $f$. Let $\{\lambda_1, \dots,\lambda_k\}$ be its distinct roots. Let $v_i$ be corresponding eigenvectors, i.e. $f(v_i)=\lambda_i v_i$. Then $v_i$ are also eigenvectors of $f^*$ i.e. $f^*(v_i)=\overline{\lambda_i} v_i$. Let's take $W=\langle v_1,\dots,v_k\rangle $ and $V=W\oplus W^{\perp}$. And I guess we have to do smth with $W^{\perp}$.
Would be very thankful is someone can show how to prove this direction, please?
Best Answer
One proof is as follows:
Suppose that any eigenvector of $f$ is also an eigenvector of $f^*$. It is clear that for any eigenvector $x$ of $f$, we have $f(f^*(x)) = f^*(f(x))$. With that, we can see that $f(f^*(x)) = f^*(f(x))$ for any $x$ that can be written as a linear combination of eigenvectors of $f$. So, if the eigenvectors of $f$ span $V$ (i.e. if $f$ is diagonalizable), then $f$ must be normal.
We prove that $f$ is diagonalizable inductively on $n = \dim(V)$; the $n = 1$ case is trivial. Let $\lambda,v$ be any eigenvalue/eigenvector pair. We have $f^*(v) = \mu v$ for some $\mu \in \Bbb C$. We note that for any $w \in \langle v \rangle^\perp$, we have $$ (f(w),v) = (w,f^*(v)) = (w,\mu v) = \bar \mu (w,v) = 0. $$ Thus, $f(\langle v \rangle^\perp)\subset \langle v \rangle^\perp$. Now, the restriction $g = f|_{\langle v \rangle^\perp}$ is such that any eigenvector of $g$ is also an eigenvector of $g^*$. By the inductive hypothesis, $g$ must be diagonalizable. We thus conclude that $f$ is diagonalizable, as was desired.
The originally posted proof:
Suppose that any eigenvector of $f$ is also an eigenvector of $f^*$. Note that if $x\neq 0$ is such that $f(x) = \lambda x$ and $g(x) = \mu x$, then we have $$ \lambda (x,x) = (\lambda x,x) = (f(x),x)=(x,f^*(x)) = (x,\mu x) = \bar \mu (x,x), $$ which implies that $\mu = \bar \lambda$. That is, if $x$ is an eigenvector of $f$ associated with $\lambda$, then it is an eigenvector of $f^*$ associated with $\bar \lambda$.
It is clear that for any eigenvector $x$ of $f$, we have $f(f^*(x)) = f^*(f(x))$. With that, we can see that $f(f^*(x)) = f^*(f(x))$ for any $x$ that can be written as a linear combination of eigenvectors of $f$. So, if the eigenvectors of $f$ span $V$ (i.e. if $f$ is diagonalizable), then $f$ must be normal.
So, suppose for the sake of contradiction that $f$ is not diagonalizable. Then there exists a vector $x$ and eigenvalue $\lambda$ for which $(f-\lambda \operatorname{id})(x) \neq 0$ and $(f-\lambda \operatorname{id})^2(x) = 0$. Let $g = f - \lambda I$; note that every eigenvector of $g$ is also an eigenvector of $g^*$. We see that $g(x) \neq 0$, and $g^2(x) = 0$. Thus, $y = g(x)$ is an eigenvector of $g$ associated with $\lambda = 0$. It follows that $g^*(y) = \bar 0 \cdot y = 0$. In other words, we have $g^*(g(x)) = 0$. However, $x \in \ker g^*g = \ker g$, which means that $g(x) = 0$, which is a contradiction.