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Let $V$ be a Hermitian finite-dimensional vector space and $f$ is an operator on $V$ and if $W$ is $f$-invariant then $W^{\perp}$ is $f$-invariant. Prove that $f$ is normal operator.

Proof: Let's prove by induction on $\dim V$.

If $\dim V=1$, i.e. $V=\langle e_1\rangle$ then $f(e_1)=\lambda e_1$. Then $f^*(e_1)=\mu e_1$ and it's trivial to check that $ff^*=f^*f$, i.e. $f$ is normal. Question: Am I right that in this case I do not use anything from the problem statement?

Suppose it is true for all Hermitian spaces of dimension $\leq n-1$.

Let $\dim V=n$. Since the ground field is $\mathbb{C}$ then there is eigenvector $v$ associated with eigenvalue $\lambda$. Let $W=\langle v\rangle $, then $\dim W^{\perp}=n-1$. Since $W$ is $f$-invariant then $W^{\perp}$ is $f$-invariant and I can consider restriction $h:=f|_{W^{\perp}}$ where $h:W^{\perp}\to W^{\perp}$ is operator. Our goal to apply induction hypothesis to operator $h$. But in order to do that we have to show that $h$ has the desired property: if $U$ is $h$-invariant then $U^{\perp}$ is $h$-invariant (here by $U^{\perp}$ I mean orthogonal complement relative to $W^{\perp}$).

Let $U$ is $h$-invariant subspace of $W^{\perp}$ then $h(U)=f(U)\subseteq U$ which shows that $U$ is $f$-invariant subspace of $V$ then it follows that $U^{\perp}$ is $f$-invariant, but by $U^{\perp}$ I mean orthogonal complement relative to $V$. Then I am got confused completely.

Can anyone show how to make this reasoning more clearer, please?

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One approach is as follows: not that $W^\perp$ is $f$ invariant if and only if $W$ is $f^*$ invariant. By considering one dimensional subspaces $W$, we see that every eigenvector of $f$ is an eigenvector of $f^*$. By applying the result from your previous post, we conclude that $f$ must be normal.


Regarding your attempted proof:

We have to show that $h$ has the desired property: if $U$ is $h$-invariant then $U^{\perp}$ is $h$-invariant (here by $U^{\perp}$ I mean orthogonal complement relative to $W^{\perp}$).

In other words: if $U$ is $h$ invariant, then we need to show that $U^\perp \cap W^\perp$ is $h$ invariant. Indeed: if $U$ is $h$ is invariant, then it is $f$ invariant. Thus, both $U^\perp$ and $W^\perp$ are $f$-invariant. The intersection of $f$-invariant spaces is $f$-invariant, so $U^\perp \cap W^\perp$ is $f$-invariant. So, $U^\perp \cap W^\perp$ (i.e. the orthogonal complement of $U$ relative to $W^\perp$) is $h$-invariant, as was desired.

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  • $\begingroup$ I haven’t looked at your approach yet; I’ll do so when I find the time tomorrow $\endgroup$ Commented Apr 8, 2020 at 4:36
  • $\begingroup$ Thanks a lot! BTW very elegant solution! +1 $\endgroup$
    – RFZ
    Commented Apr 8, 2020 at 14:57
  • $\begingroup$ @ZFR You're welcome. See my latest edit. $\endgroup$ Commented Apr 8, 2020 at 15:45
  • $\begingroup$ Please sorry me that I am asking you for help! But could you take a look at this, please? math.stackexchange.com/questions/3616578/… $\endgroup$
    – RFZ
    Commented Apr 9, 2020 at 2:15
  • $\begingroup$ Regarding your latest edit: you said that by $U^{\perp}$ you mean orthogonal complement relative to $W^{\perp}$. And then you are proving that $U^{\perp}\cap W^{\perp}$ is $h$ invariant. But $U^{\perp}\cap W^{\perp}=U^{\perp}$, right? $\endgroup$
    – RFZ
    Commented Apr 9, 2020 at 2:24

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