How to show that $T|_w^* = (T|_w)^*$

Question

Let $T:V \rightarrow V $ linear operator in the inner product space $V$ such that $W$ is subspace of $V$ and for every $T(W) \subset W$ $\quad T^*(W) \subset W$

I am asked to show that $$T|_w^* = (T|_w)^*$$

so what I did so far is to show that $T(W^\perp)\in W^\perp$ and $T^*(W^\perp)\subset W^\perp$

How should I procced from here? is what I did is fine or there is no reason to do that?

Answer 1

Presumably you mean you want to show that $T^*|_W=(T|_W)^*$?

If so then what you did is unnecessary I think. You should use the fact that if $A$ is a linear operator on $V$, then $A^*$ is characterized by the relation $(Av,w)=(v,A^*w)$ for all $v,w\in V$, where $(\cdot,\cdot)$ denotes the inner product on $V$.

If that's not enough, heres the proof:

Let $v,w\in W$. Then we have $(T|_W(v),w)=(Tv,w)=(v,T^*w)=(v,T^*|_W(w))$. But we also have $(T|_W(v),w) = (v,(T|_W)^*(w))$ by definition. Since this is true for all $v,w\in W$, we must have $T^*|_W=(T|_W)^*.$