Restriction of adjoint operator to invariant subspace

Suppose that $$V$$ is a finite dimensional euclidean space and let $$f:V\to V$$ is an operator. Let $$f^*:V\to V$$ is an adjoint operator. Let $$W$$ is a subspace of $$V$$ which is invariant under $$f$$ and $$f^*$$. Then we can define new operators $$(f|_W)^*:W\to W$$ and $$f^*|_W:W\to W$$.

So probably I will ask a stupid question: how to show that $$(f|_W)^*=f^{*}|_W$$? Since domain of those operators is $$W$$, then we need to show that for each $$x\in W$$ we have $$(f|_W)^*(x)=f^{*}|_W(x)$$.

The RHS of above is $$f^{*}|_W(x)=f^{*}(x)$$ by definition of restriction. I want to show that LHS if also equal to $$f^{*}(x)$$, i.e. $$(f|_W)^*(x)=f^{*}(x)$$. Intuitively I know that this is true but cannot persuade me that this is formally correct.

Would be very thankful for any help and comments!

2 Answers

Fix $$x \in W$$. Then, for any $$y \in W$$, \begin{align} \langle (f|_W)^*(x),y \rangle &= \langle x,(f|_W)(y) \rangle \\ &= \langle x,f(y) \rangle \\ &= \langle f^*(x),y \rangle \end{align} Hence, $$\langle (f|_W)^*(x) - f^*(x),y \rangle = \langle (f|_W)^*(x),y \rangle - \langle f^*(x),y \rangle = 0.$$ Since the latter works for any $$y \in W$$, for $$y = (f|_W)^*(x) - f^*(x)$$ we have that $$\langle (f|_W)^*(x) - f^*(x), (f|_W)^*(x) - f^*(x) \rangle = 0$$ but this only happens when $$(f|_W)^*(x) - f^*(x) = 0$$, i.e. $$(f|_W)^*(x) = f^*(x)$$.

• Thanks a lot for your detailed reply! But what is wrong with my reasoning? Is it wrong?
– RFZ
Commented Mar 19, 2020 at 1:04

I'm assuming by "euclidean space" you mean "inner product space", since otherwise I'm not sure how to interpret "adjoint operator". It's likely I'm just unaware of alternate terminology!

You can prove $$f^*|_W - (f|_W)^* = 0$$ as follows: let $$x,y \in W$$ be arbitrary. Then

$$\newcommand{\angles}[1]{\left\langle#1\right\rangle}\angles{(f^*|_W - (f|_W)^*)(x), y} = \angles{f^*(x),y} - \angles{(f|_W)^*(x),y} = \angles{x,f(y)} - \angles{x, f|_W(y)} = 0.$$

By non-degeneracy of the inner product, $$(f^*|_W - (f|_W)^*)(x) = 0$$ for all $$x \in W$$, as desired.

Note that if these two maps are going to be equal in this level of generality, it ought to be the case that we can prove they're equal using only the definition of "adjoint map". Basically, this proof says that both maps are the adjoint of $$f|_W$$, so they must be the same.