Find a matrix $N$ s.t. $N^2=H$

Question

Given the below matrix, find a matrix $N$ s.t. $N^2=H$

$$H=\begin{bmatrix}2&i&2&1+i\\-i&2&-2i&1-i\\2&2i&5&2+2i\\1-i&1+i&2-2i&3\end{bmatrix}$$

I solved for the characteristic polynomial, which is $(x-1)^3(x-9)$ Any help will be appreciated.

Answer 1

$H$ is Hermitian, so it is diagonalizable: $H=U\Sigma U^*$. You've found the eigenvalues are non-negative, so define $N:=U\sqrt{\Sigma}U^*$ (where the square root applies to the diagonal entries, i.e. eigenvalues of $H$). Then $N^2=H$.

Answer 2

Your matrix is diagonalizable. More precisely, if$$M=\begin{pmatrix}1+i & -1-i & -2 & -i \\ 1-i & 0 & 0 & 1 \\ 2+2 i & 0 & 1 & 0 \\ 2 & 1 & 0 & 0\end{pmatrix},$$then$$M^{-1}.H.M=\begin{pmatrix}9&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}.$$So, an answer to your question is$$M.\begin{pmatrix}3&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}.M^{-1}$$