If $A$ is a symmetric matrix. Then all the eigenvalues of $A^2$ are nonnegative For the answer, we can get support from two claims:
Claim 1: If $A$ is real and symmetric, then it has real eigenvalues. Proof here
Claim 2: If $\lambda$ is an eigenvalue of $A$, then $\lambda^2$ is an Eigenvalue of $A^2$. Proof here 
So by those two claims we can say that if $\lambda$ is an eigenvalue of $A$, then the corresponding eigenvalue $\lambda^2$ of $A^2$ is nonnegative. But my problem is how can we guarantee that it will cover all the possible eigenvalues of $A^2$? 
 A: Begin with the diagonalization
$$
A=PDP^{-1}
$$
where $P$ is the matrix of eigenvectors and $D$ is the diagonal matrix with eigenvalues on the main diagonal. Then
$$
A^2=(PDP^{-1})(PDP^{-1})=PD^2P^{-1}
$$
from which the claim follows.
A: A symmetric matrix is diagonalizable,  $D=PAP^{-1}$ where $A$ is symmetric and $D$ diagonal. $D^2=PA^2P^{-1}$, this implies that the eigenvalues of $A^2$ are the square of the eigenvalues of $A$.
A: More than this is true: if $B$ is any real matrix, then the eigenvalues of $B^TB$ are real and nonnegative.
The fact that the eigenvalues are real follows from your claim 1.
Suppose $\lambda$ is an eigenvalue with eigenvector $v$; then
$$
\lambda(v^Tv)=v^T(\lambda v)=v^TB^TBv=(Bv)^T(Bv)\ge0
$$
so $\lambda\ge0$.
In your case $A$ is symmetric, so $A^2=A^TA$.
With diagonalization you can indeed prove that the eigenvalues of $A^2$ are the squares of the eigenvalues of $A$, but it is not necessary for proving the statement you have.
A: Suppose $J$ is a Jordan form of $A$, then it should be clear that the diagonal elements of $J^2$ are the squares of the diagonal elements of $J$. In other words, the eigenvalues of
$A^2$ are exactly the squares of the eigenvalues of $A$. This is true for any
square matrix, and is true much more generally (that is, the eigenvalues of the function of a matrix are the functions of the eigenvalues of the matrix).
Hence the eigenvalues are all of the form $x^2$ where $x$ is real.
A: Let $\lambda = a+ib\in \mathbb{C}\setminus \mathbb{R}$. For any $x\in\mathbb{C}^n$ we have
\begin{align}
\|(A^2-\lambda I)x\|^2\|x\|^2 &\ge \left|\langle (A^2-\lambda I)x,x \rangle\right|^2 \\
&= \left|\langle A^2x,x\rangle - \lambda\langle x,x\rangle\right|^2\\
&= \left|\left(\|Ax\|^2 - a\|x\|^2\right) + ib\|x\|^2\right|^2\\
&= \left(\|Ax\|^2 - a\|x\|^2\right)^2 + |b|^2\|x\|^2\\
&\ge |b|^2\|x\|^4
\end{align}
so $\|(A^2-\lambda I)x\| \ge |b|\|x\|$. Since $b\ne 0$, we see that $A^2-\lambda I$ is bounded from below and hence injective. Therefore $\lambda$ is not an eigenvalue of $A^2$.
Now let $\lambda < 0$. For any $x\in\mathbb{C}^n$ we have
$$\|(A^2-\lambda I)x\|\|x\| \ge \langle (A^2-\lambda I)x,x\rangle = \langle A^2x,x\rangle - \lambda\langle x,x\rangle = \|Ax\|^2 - \lambda \|x\|^2 \ge -\lambda \|x\|^2$$
so again $A^2-\lambda I$ is bounded from below and hence injective. Therefore $\lambda$ is not an eigenvalue of $A^2$.
We conclude that the eigenvalues of $A^2$ are contained in $[0, +\infty\rangle$.
