I am trying to show that matrix $A$ is symmetric positive semidefinite if and only if there exists a symmetric matrix $B$ such that $B^2 = A$.
Here is my solution, any comments?
I have attempted to solve this with diagonalization of A and using the determinants of A to show there is a matrix $B^2$ that equals $A$. If there is a matrix $B$ such that $B = P \operatorname{diag}(\lambda^{1/2}...\lambda_n^{1/2})P^{-1}$ Let matrix $ D = \operatorname{diag}(\lambda_1^{1/2}...\lambda_n^{1/2})$ if $ PBP^{-1} = D$ then $\det|D| = \det|P|\det|B|\det|P^{-1}$ where $\det|P| \cdot 1/\det|P| = 1$, therefore there exists a matrix $B^2$ where $\det|A| = \det|B| \cdot \det|B|$ and this matrix $B^2 = A$.
To extrapolate further: $PAP^{-1} = D^2$
$$ PAP^{-1} = \operatorname{diag}(\lambda^{1/2}...\lambda_n^{1/2}) = \operatorname{diag}(\lambda^{1/2}...\lambda_n^{1/2}) \cdot \operatorname{diag}(\lambda^{1/2}...\lambda_n^{1/2}) = PBP^{-1} PBP^{-1}$$
$$PB(P^{-1}P)BP^{-1} = PB^2P^{-1} = PAP^{-1}$$ reduce to $B^2 = A$
You seem to have (more or less) correctly shown that if $A$ is positive semidefinite, then there is a $B$ such that $A = B^2$.
Now, suppose that $B$ is symmetric, and $A = B^2$. We note that $\lambda$ is an eigenvalue of $A$ if and only if $\lambda = \mu^2$ for some eigenvalue $\mu$ of $B$. The eigenvalues of $B$ are real. Thus, the eigenvalues of $A$ are non-negative. Moreover, $A$ is symmetric since $B$ is symmetric.
Since $A$ is symmetric with non-negative eigenvalues, it is positive semidefinite, as desired.