Show that a matrix $A$ is symmetric positive semidefinite iff there exists a symmetric matrix $B$ such that $B^2=A$

Question

There is a hint here that says to use the diagonal form of the matrix $A$.

Answer 1

If $A$ is positive semidefinite, then $A=V\Lambda V^*$, where $\Lambda$ is diagonal and positive semidefinite, and $V$ is unitary. Then, if $\sqrt\Lambda$ is the diagonal matrix with the square roots of the elements of $\Lambda$, then $B=V\sqrt\Lambda V^*$, and $B^2=A$.

On the other hand, if $A=B^2$ and $B$ is symmetric, then $B=V\Lambda V^*$, and $A=B^2=V\Lambda V^*V\Lambda V^*=V\Lambda^2 V^*$. Since $B$ is symmetric, $\Lambda$ is real, and then $\Lambda^2$ is positive semidefinite, so that $A$ is positive semidefinite.