Let $A=\begin{bmatrix}1 & 1+j \\1-j & 2 \end{bmatrix}$. Find $A^{\frac{1}{2}}$ and a $B\in\mathbb{C}^2$ such that $A=BB^*.$
I have tried to write down three equations using elements of $B$ and one equation using the fact that determinant of $A$ is zero. But it didn't help.
Any smarter way to solve this problem?
Hint: If $u$ is a unit-vector spanning the image of $A$, then we can take $B = \alpha u$ for some scalar $\alpha > 0$.
We then have $$ A = BB^* = \alpha^2 uu^*. $$ If we take $M = \alpha \,uu^*$, then we find that $$ M^2 = (\alpha uu^*)(\alpha uu^*) = \alpha^2 u(u^*u)u = \alpha^2 uu^* = A. $$ Because $M$ is positive definite, we conclude that $M = A^{1/2}$.