Prove that a Linear Transformation is Normal, such that $T^2=T$

Question

Let V be a finite inner product space and $T:V\rightarrow V$ a linear transformation such that $$T^2=\frac{1}{2}(T+T^*)$$.

Prove that

1. $T$ is a "Normal Transformation", such that $$TT^*=T^*T$$
2. $T^2=T$

Proving the first statement isn't hard and I did it that way:

$T^*=2T^2-T$ $\Rightarrow TT^*=T(2T^2-T)=(2T^2-T)T=T^*T$

Therefore, $TT^*=T^*T$.

However, proving the second statement was much harder, and I couldn't figure it out.

Thanks,

Alan

Answer 1

Once you know that $T$ is normal, then it has a basis of eigenvectors. An eigenvalue $\lambda$ must satisfy $$ \lambda^{2}=\frac{1}{2}(\lambda+\overline{\lambda})=\Re\lambda. $$ The only way $\lambda^{2}$ is real is if $\lambda$ is real or is purely imaginary. However, $\lambda=ir$ for some real $r$ gives $-r^{2}=\Re(ir)=0$. So $\lambda$ must be real. Furthermore $\lambda^{2}=\Re\lambda=\lambda$, which means that $T^{2}=T$ also holds.