Show that $A^2=A$ suppose that A is normal and $A^5=A^4$

Question

In the beginning this seems easy to me - just multiply $A^{-3}$ on both sides, but I realized that $A$ and its associated diagonal matrix $\mathbf{\Lambda}$ are not nessarily invertible. So what else can I try? Any help will be appreciated!

Answer 1

$A$ is normal $\implies$ $A$ is unitarily diagonalizable, i.e. $A=P^*\Lambda P$ for the diagonal matrix $\Lambda$ of eigenvalues of $A$ and a unitary matrix $P$. This is actually an if and only if statement.

Answer 2

I'm answering this question myself as @lhf suggested and see if I am correct...

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Since $A$ is normal, by Spectral Theorem for Normal Matrices, there exists some diagonal matrix $D$ such that $A=UDU^*$. So we have

$$ A^{5} = (UDU^*)^{5} = \underbrace{(UDU^*)\cdots(UDU^*)}_{5 \text{ times}} = UD\ (UU^*)\ D\ (U\cdots U^*)\ DU^*=UD^{5}U^*$$

and similarly for $A^{4}$. Then

$$UD^{5}U^* = UD^{4}U^*$$

this implies for $D=\text{diag}(\lambda_1,\ldots,\lambda_n), \lambda_i=0\text{ or }1 \text{ for } i\in[n].$ Therefore $$A^2=UD^{2}U^*=UDU^*=A.$$

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Moreover, we can show that $A$ is actually Hermitian:

$$\begin{aligned} A^*&=(UDU^*)^* \\ &=(U^*)^*D^*U^*\\ &=UDU^* \text{ as D is real from above}\\ &= A \end{aligned}$$