A question from Golan's linear algebra: A question from Golan's linear algebra:
Let $A\in M(n,\mathbb R)$ (which denotes the set of all $n\times n$ matrices, for some $n\geq 2$) be symmetric. Does there exist a symmetric matrix $B$ such that $B^2=A$?
It asks again whether it is possible to find such a matrix in case $A$ is symmetric and positive definite? How to do this? Any hints to approach them. I am totally clueless

Hint: Every symmetric matrix is unitarily diagonalizable by the spectral theorem.  How could you (easily) construct a square root of a diagonal matrix?
If $B^2 = A$, what could the eigenvalues of $B$ be?  If $A$ is symmetric, what do we know about its eigenvalues?
The answer will be no in general, but yes if $A$ is also positive definite.

If $A = UDU^*$ where $D$ is diagonal, try
$$
B = U \pmatrix{\sqrt{\lambda_1}\\&\ddots \\ && \sqrt\lambda_n}U^*
$$
