# Thread: T is a projection and normal, then T is orthogonal projection

1. ## T is a projection and normal, then T is orthogonal projection

Question: Let $\displaystyle T$ be a normal operator on a finite-dimensional inner product space. Prove that if $\displaystyle T$ is a projection, then $\displaystyle T$ is also an orthogonal projection.

Attempt: Since $\displaystyle T$ is normal, $\displaystyle TT^* = T^*T$ and since $\displaystyle T$ is a projection, then $\displaystyle T=T^2$.

Since $\displaystyle T$ is an orthogonal projection if and only if $\displaystyle T^2 = T = T^*$, we already have the left equality by that fact that $\displaystyle T$ is a projection. We only need to show that $\displaystyle T=T^*$ and we'll be done.

Somehow, I figure that I can do this by showing that $\displaystyle <T(x),y>=<x,T(y)>$ but how?

2. Since $\displaystyle T$ is normal, it is unitarily equivalent to a diagonal matrix. Since $\displaystyle T=T^2$, its eigenvalues can be only $\displaystyle \pm 1$. Thus $\displaystyle T=UDU^*$ and $\displaystyle T^*=U^{**}D^*U^*=UDU^*=T$.