$A\in M_{n}(C)$ and $A^*=-A$ and $A^4=I$ Let $A\in M_{n}(C)$ be a matrix such that $A^*=-A$ and $A^4=I$.
I need to prove that the eigenvalues of A are between $-i$ to $i$ and that $A^2+I=0$
I didn't get to any smart conclusion.
Thanks

Do you recall that hermitian matrices ($A^*=A$) must have real eigenvalues?  Similiarly, skew-hermitian matrices, i.e. $A^*=-A$, must have pure imaginary eigenvalues.
(see Why are all nonzero eigenvalues of the skew-symmetric matrices pure imaginary?)
Also, since $A$ is skew-hermitian, then $A$ is normal too, i.e. $A^*A=AA^*$, so we can apply the spectral theorem:  there exists a unitary matrix $U$ such that $A=UDU^{-1}$, where $D$ is a diagonal matrix, whose diagonal entries are the eigenvalues of $A$.
Thus we know that $A^4=(UDU^{-1})^4=UD^4U^{-1}=I$, so $D^4=I$, so all the eigenvalues are 4th roots of unity, i.e. $1,-1,i,\text{ or} -i$.  But we already know the eigenvalues are pure imaginary, so all the eigenvalues are $i$ or $-i$.
 So $D^2=-I$.
Finally, we have $A^2=(UDU^{-1})^2=UD^2U^{-1}=U(-I)U^{-1}=-I$, i.e. $A^2+I=0$.
