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

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.

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?