If $A + I$, $A^2 + I$ and $A^3 + I$ are all unitary, show that $A$ is the zero matrix Let $A$ be an $n \times n$ 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 know eigenvalues of a unitary matrix has modulus 1, so if $\lambda$ is an eigenvalue of A, then the eigenvalues of $A + I$, $A^2 + I$, $A^3 + I$ are $ \lambda + 1, \lambda^2 + 1, \lambda^3 + 1$ all on the circle $|z+1|<1$. 
Where to go from here though?
 A: If $X+I$ is unitary, then $(X+I)(X+I)^*=XX^*+X+X^*+I=I$, so $XX^*=-(X+X^*)$.  Similarly, $X^*X=-(X+X^*)$, so $XX^*=X^*X$.  Since normal matrices are diagonalizable, it suffices to show that the only eigenvalue is zero.
Let $\lambda$ be an eigenvalue.  Then $$(1+\lambda)(1+\overline{\lambda})=(1+\lambda^2)(1+\overline{\lambda^2})=(1+\lambda^3)(1+\overline{\lambda^3})=1.$$  Expanding, $|\lambda|^2=-2\operatorname{Re}(\lambda)$, $|\lambda|^4=-2\operatorname{Re}(\lambda^2)$, and $|\lambda|^6=-2\operatorname{Re}(\lambda^3)$.  In particular, $\lambda, \lambda^2$, and $\lambda^3$ all have non-positive real part.  If $\lambda\neq 0$, then we can make deductions about the argument $\theta$ of $\lambda$.  We need $\theta \in [\pi/2,3\pi/2]$ to satisfy the first equation, $\theta\in [\pi/4,3\pi/4]\cup [5\pi/4,7\pi/4]$ to satisfy the second, and $\theta \in [\pi/6,3\pi/6]\cup [5\pi/6,7\pi/6] \cup [9\pi/6,11\pi,6]$ to satisfy the third.  However, the only solution this system is for $\lambda$ to be pure imaginary, but then going back to the equation $|\lambda|^2=-2\operatorname{Re}(\lambda)=0$, we have $\lambda=0$.

An alternative approach is to deal directly with the equations $X^*X=-(X+X^*)$ for $X=A,A^*, A^2, A^3$.  From the first two, we deduce normality of $A$ (as above).  From the first and the third (combined with normality), we deduce $A^2+(A^*)^2=A+A^*$, and from the first and the fourth (combined with normality) we deduce $A^2A^*+A(A^*)^2=0$.  Factoring out $A^* A$ from this last equation and using what we know, we have $(A+A^*)^2=0$.  Expanding and using $A^2+(A^*)^2=-A^*A=A+A^*$, we have $A+A^*=0$, so $A$ is skew-hermetian.  
Going back to our original equations ($X=A$), $A$ being skew-hermetian means $A^*A=0$.  However, the entries of $A^*A$ are the various inner products of the columns of $A$.  In particular, each column has norm $0$, and is zero, hence $A=0$.
