Proving that a matrix has positive eigenvalues. Let's say that $\mathrm{A}$ is a diagonalizable matrix with distinct nonzero
eigenvalues. Prove that $\mathrm{A}^2$ has positive eigenvalues.
Ok I have a general idea how to do this, but not sure about how formal I can make it look.
Since by definition, if matrix $\mathrm{A}$ is diagonalizable, then it can be written as  $\mathrm{A}=\mathrm{P}\mathrm{D}\mathrm{P}^{-1}$.
With invertible matrix $\mathrm{P}$ and Diagonal Matrix $\mathrm{D}$.  The Matrix $\mathrm{P}$ is formed with the eigenspaces of the respective eigenvalues of $\mathrm{A}$, while Matrix $\mathrm{D}$ is formed WITH the EIGENVALUES of Matrix $\mathrm{A}$ directly. So if we square both sides of the equation, the diagonal contents of $\mathrm{D}$ would be pretty much guaranteed to be non negative because two numbers of the same sign always yields a positive. How am I gonna prove it?
 A: You don't need diagonalizable, only that the eigenvalues of $A$ are real. The eigenvalues of $A^2$ are the squares of the eigenvalues of $A$ (this can be seen using the Jordan form). So, if the eigenvalues of $A$ are real and nonzero, their squares are positive. 
A: Hint: What happens to the entries of a diagonal matrix when you square it? What is $(PDP^{-1})^2,$ expanded and simplified?
A: Schur decomposition: since $A=QRQ^*$ with $Q$ unitary and $R$ upper triangular with the eigenvalues of $A$ on the diagonal of $R$, we have $A^2=(QRQ^*)^2=QR^2Q^*$. The matrix $R^2$ is upper triangular as well with the diagonal elements equal to the squares of the diagonal elements of $R$; hence the eigenvalues of $A^2$ are squares of the eigenvalues of $A$.
If $A$ has real eigenvalues, then their squares are real as well and non-negative.
A: It can also follow from the Spectral Mapping Theorem. Simple idea is:
$\begin{align}
Ax &= \lambda x \\
A^2x &= \lambda Ax = \lambda^2 x
\end{align}$
Therefore, if $\lambda$ is an eigenvalue of $A$, then $\lambda^2$ is an eigenvalue of $A^2$ with the same eigenvector. This can also be generalized to polynomials $p(A)$ (actually this is what Spectral Mapping Theorem states).
