Prove that A is zero matrix

Question

Let $A$ be an $n×n $ complex matrix such that the three matrices $A+I$ , $A^2+I $ , $ A^3+I$ are all unitary .Prove that$ A$ is the zero matrix

I try to show that

$Trace( A^{\theta}A) =0$ where $A^{\theta }$ is conjugate transpose of matrix $A$

$\because $ $Trace( A^{\theta}A)$ = $|a_{11}|^2 + |a_{12}|^2....|a_{nn}|^2$

$A+I$ is unitary ,so

$(A+I)^{\theta}(A+I)= I $

$\implies (A^ {\theta}+I)(A +I) =I $

$A^ {\theta}A+ A^ {\theta}+A = 0$

$ Trace( A^{\theta}A)= -( Trace( A^{\theta}+A))$ $ \implies Trace( A^{\theta}A)=-2$( sum of real parts of each diagonal entry of A

I don't know how to proceed further Please help

Answer 1

Note that the eigenvalues of $A^k + I$ are of the form $\lambda^k + 1$ for all eigenvalues $\lambda$ of $A$. On the other hand, the eigenvalues of a unitary matrix must have absolute value $1$. Thus, each eigenvalue $\lambda$ of $A$ satisfies $|\lambda^k + 1| = 1$ for $k = 1,2,3$.

I claim that the only $\lambda$ for which this holds is $\lambda = 0$. Thus, $A$ has $0$ as its only eigenvalue.

On the other hand, because $A + I$ is unitary, $A^\theta A$ is normal (in particular, we find that $A^\theta A = AA^\theta = -A - A^\theta$). By the spectral theorem, $A$ is unitarily diagonalizable. Because $A$ is diagonalizable with $0$ as its only eigenvalue, $A = 0$.

Answer 2

Since $A+I$ is unitary, $A$ is normal. Thus it is sufficient to prove that the only eigenvalue of $A$ is $0$.

Since $A+I$ is unitary, an eigenvalue of $A$ must be of the form $z-1$ where $|z|=1$. The corresponding eigenvalues of $A^2+I$ and $A^3+I$ are then $(z-1)^2+1$ and $(z-1)^3+1$. It can be seen that the only $z$ with $|z|=1$ and $|(z-1)^3+1|=1$ is $1$.