Let $T$ be a linear operator such that it is both self-adjoint and unitary

Question

Let $T$ be a linear operator on a finite dimensional inner product space $V$ such that it is both self-adjoint and unitary.Then prove that it can only arise from the subspace $W$ s.t.

$$V=W \bigoplus W^{\perp}$$

and if $a=a_1+a_2$ then $T(a)=a_1-a_2$.

Now if $T$ is self-adjoint then $T=T^*$ for an orthodontist basis of $V$.Also, $TT^*=T^*T=I$ and $T$ carries orthonormal basis vectors to orthonormal basis vectors if $T^*$ is unitary.

now from here we can conclude that $$T^2=I$$ .Hence $T$ is an idempotent linear operator. since $T$ is an idempotent linear operator then we see that,

$$V = R(T) \bigoplus N(T)$$ Now we call R(T) to be as $W$.

We need to show that $N(T)=R(T)^{\perp}$ .

let $x \in N(T)$.Then, we take a vector $y \in R(T)$ ,we see that

$$\langle y, T^*T(x)\rangle=\langle y, T^*(0)\rangle=\langle y, 0\rangle=0\,.$$

Then, it is easy to
conclude that $N(T)=W^{\perp}$.

Now how do I proceed from here.

This from Hoffman exercise -8.4 no-12

I would prefer some hints and intuitions behind those hints instead of complete answers.

Answer 1

Hint

The minimal polynomial of $T$ divides $p(x) = x^2-1$. $T$ is diagonalizable. What are the possible values for the eingenvalues? What happens if you consider an eigenvectors basis?