# $f(W)\subset W, f(W^{\perp})\subset W^{\perp}$ implies $f$ normal operator

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?

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.

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.

• I haven’t looked at your approach yet; I’ll do so when I find the time tomorrow Apr 8, 2020 at 4:36
• Thanks a lot! BTW very elegant solution! +1
– RFZ
Apr 8, 2020 at 14:57
• @ZFR You're welcome. See my latest edit. Apr 8, 2020 at 15:45
• Please sorry me that I am asking you for help! But could you take a look at this, please? math.stackexchange.com/questions/3616578/…
– RFZ
Apr 9, 2020 at 2:15
• 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?
– RFZ
Apr 9, 2020 at 2:24