If $A$ is a normal matrix such that $A^{247} = A^{246}$, how to prove that $A$ is an orthogonal projection?

Question

Let $A$ be a normal matrix and $A^{247} = A^{246}$.

Then I want to prove:

1. $A^2 = A$.
2. $A$ is Hermitian.

I'm not sure on how to prove this. If $A$ is normal, then $A$ is unitarily diagonalizable. Then $A^{247}=A^{246}$ implies $U^\ast A^{247}U = U^\ast A^{246}U$. Then if $D = U^\ast AU$, can I say that $D^{247}=D^{246}$ ? I don't know how to continue or if my thinking was correct.

Please help.

Answer 1

You have the right idea. Regarding your specific question about $D$, $$D^{247} = (U^* A U)^{247} = U^* A^{247} U = U^* A^{246} U = (U^* A U)^{246} = D^{246}.$$

Answer 2

Write, $A = U \Lambda U^\dagger$, where $U^\dagger$ is the conjugate transpose of $U$ and $\Lambda = \operatorname{diag}\{\lambda_1, \lambda_2, \dots \lambda_n\}$.

Now, $A^{247} = U \Lambda^{247} U^\dagger = U \operatorname{diag}\{\lambda_1^{247}, \lambda_2^{247}, \dots \lambda_n^{247}\} U^\dagger = U \operatorname{diag}\{\lambda_1^{246}, \lambda_2^{246}, \dots \lambda_n^{246}\} U^\dagger$.

Clearly $\lambda_i^{247} = \lambda_i^{246} \Rightarrow \lambda_i = 0$ or $1$ for all $i$.

Now, $A^2 = U \operatorname{diag}\{\lambda_1^2, \lambda_2^2, \dots \lambda_n^2\} U^\dagger = A$.

Also, $A^\dagger =A$.

Answer 3

Suppose $A$ is normal. Then $A^2x=0$ iff $Ax=0$. Therefore $A^{246}(A-I)=0$ implies $A(A-I)=0$, and that makes $A$ a projection. Because $\mathcal{N}(A)=\mathcal{N}(A^*)$ for a normal $A$, then $A(A-I)=0$ implies $A^*(A-I)=0$ and, hence $A^*=A^*A$, which also gives $A=(A^*A)^*=A^*A=A^*$. Hence, $A$ is Hermitian.