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

**Last_Singularity** · December 2nd 2009, 09:27 PM

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 $<T(x),y>=<x,T(y)>$ but how?

**Bruno J.** · December 2nd 2009, 10:14 PM

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 $T^*=U^{**}D^*U^*=UDU^*=T$ .

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

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