<s>[INST] Find a matrix $B$ such that $B^2=A$ $A= \begin{pmatrix} 1 & 2+3i \\ 2-3i & -1 \end{pmatrix}$
What is matrix $B$ such that $B^2=A$?  
Its eigenvalues are $\sqrt{14}, -\sqrt{14}$
and I tried to use formula $B=U\sqrt{\lambda}U^{H}$ where $U$ is unitary matrix.
But then $\sqrt{　}$ of $-\sqrt{14}$ is not possible.  
How can I solve this?
 [/INST] The characteristic polynomial of $A$ is $\lambda^2 - 14$, so $A^2 - 14 I = 0$.
Thus $(a A + bI)^2 = (14 a^2 + b^2) I + 2 a b A$.  To make this equal to $A$, we want 
$2ab = 1$ and $14 a^2 + b^2 = 0$.  Thus $b = 1/(2a)$ and $14 a^2 = -1/(4a^2)$, or 
$a^4 = -1/56$. Take $a$ to be any of the four fourth roots of $-1/56$, and you have a solution. 
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