Prove that A is zero matrix 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
 A: 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, $A = 0$.
A: 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 $1$.
