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

• December 2nd 2009, 10:27 PM
Last_Singularity
T is a projection and normal, then T is orthogonal projection
Question: Let $T$ be a normal operator on a finite-dimensional inner product space. Prove that if $T$ is a projection, then $T$ is also an orthogonal projection.

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

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

Somehow, I figure that I can do this by showing that $=$ but how?
• December 2nd 2009, 11:14 PM
Bruno J.
Since $T$ is normal, it is unitarily equivalent to a diagonal matrix. Since $T=T^2$, its eigenvalues can be only $\pm 1$. Thus $T=UDU^*$ and $T^*=U^{**}D^*U^*=UDU^*=T$.