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?
Best Answer
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.