A is a Hermitian matrix ,Is a matrix U diagonizable if U(A-iI)=(A+iI)

Question

We know that $A$ is a Hermitian matrix .So $A\pm iI$ are invertible. How do I prove that $U$ is diagonal given $U(A-iI)=(A+iI).$ I tried claiming that $A\pm iI$ is Hermiatian but it is not not. I also tried with calculations, but that leads nowhere.

Answer 1

Hermitian matrices are diagonalizable so $A=PDP^{-1}$ with $D$ diagonal and $P$ invertible. Then $U(A-iI)=(A+iI)$ implies $UP(D-iI)P^{-1}=P(D+iI)P^{-1}$ which implies $P^{-1}UP(D-iI)=D+iI$. From this, if we can show $D-iI$ is invertible then: $$P^{-1}UP=(D+iI)(D-iI)^{-1}$$ which is diagonal since the inverse of a diagonal matrix is diagonal and the product of diagonal matrices is diagonal. By contradiction let us assume that $D-iI$ is not invertible, then $i$ is an eigenvalue of $D$ and thus an eigenvalue of $A$, but this implies that $A-iI$ is not invertible, which is a contradiction.

The fact that $A+iI$ is invertible is not needed to prove that $U$ is diagonalizable, however it gives you that $U$ is not only diagonalizable but invertible, since it shows that $(D+iI)(D-iI)^{-1}$ is a diagonal matrix with no zeroes in the diagonal, thus $0$ is not an eigenvalue of $U$.