Does a symmetric matrix necessarily have a symmetric square root? 
Does a symmetric matrix necessarily have a symmetric square root, and why?

If not, then does a symmetric matrix that is also semi-definite necessarily have a symmetric square root (that may or may not be semi-definite), and why?
 A: Any symmetric matrix $P$ (I assume you mean real matrices)  can be diagonalized by an orthogonal matrix $U$:
$$
P=U^T D U,\quad D=\left(\begin{array}{ccc}
\lambda_1&\ldots&0\\
\vdots&\ddots&\vdots\\
0&\ldots&\lambda_n\\
\end{array}\right),\quad U U^T=I,
$$
$\lambda_1,\ldots,\lambda_n$ are the eigenvalues of $P$, $\lambda_1,\ldots,\lambda_n\in\mathbb R$. If $P$ is positive semi-definite, then we have also $\lambda_1\ge 0,\ldots,\lambda_n\ge 0$. Consider the matrix
$$
Q=U^T \left(\begin{array}{ccc}
\sqrt{\lambda_1}&\ldots&0\\
\vdots&\ddots&\vdots\\
0&\ldots&\sqrt{\lambda_n}\\
\end{array}\right) U.
$$ 
It is easy to see that
$$
Q^2=U^T \left(\begin{array}{ccc}
\sqrt{\lambda_1}&\ldots&0\\
\vdots&\ddots&\vdots\\
0&\ldots&\sqrt{\lambda_n}\\
\end{array}\right) U
U^T \left(\begin{array}{ccc}
\sqrt{\lambda_1}&\ldots&0\\
\vdots&\ddots&\vdots\\
0&\ldots&\sqrt{\lambda_n}\\
\end{array}\right) U=
$$
$$
=U^T \left(\begin{array}{ccc}
\sqrt{\lambda_1}&\ldots&0\\
\vdots&\ddots&\vdots\\
0&\ldots&\sqrt{\lambda_n}\\
\end{array}\right) I
 \left(\begin{array}{ccc}
\sqrt{\lambda_1}&\ldots&0\\
\vdots&\ddots&\vdots\\
0&\ldots&\sqrt{\lambda_n}\\
\end{array}\right) U=
U^T \left(\begin{array}{ccc}
\lambda_1&\ldots&0\\
\vdots&\ddots&\vdots\\
0&\ldots&\lambda_n\\
\end{array}\right) U=P.
$$
Now suppose that a symmetric matrix $R$ which is not  positive semi-definite, i.e. has one or more negative eigenvalues, has a symmetric square root $S$. $S$ is diagonaliziable by an orthogonal matrix $U$,
$$
S=U^T \left(\begin{array}{ccc}
\lambda_1&\ldots&0\\
\vdots&\ddots&\vdots\\
0&\ldots&\lambda_n\\
\end{array}\right) U;
$$
$$
S^2=U^T \left(\begin{array}{ccc}
\lambda_1&\ldots&0\\
\vdots&\ddots&\vdots\\
0&\ldots&\lambda_n\\
\end{array}\right) U
U^T \left(\begin{array}{ccc}
\lambda_1&\ldots&0\\
\vdots&\ddots&\vdots\\
0&\ldots&\lambda_n\\
\end{array}\right) U=
U^T \left(\begin{array}{ccc}
\lambda_1^2&\ldots&0\\
\vdots&\ddots&\vdots\\
0&\ldots&\lambda_n^2\\
\end{array}\right) U.
$$
We can see that the eigevalues of $R=S^2$ are $\lambda_1^2\ge 0,\ldots,\lambda_n^2\ge 0$. It contradicts with the assumption that 
$R$ has negative eigenvalues.
A: Yes and no are the two answers. A symmetric matrix has a symmetric square root if and only if it is positive semi-definite. As simple proof is to use diagonalization.  
