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?

• You almost have it: what is the product of a diagonal matrix with itself? There you go....BTW, $\;P\;$ is formed with the eigenvectors of..., not with the eigenspaces. – DonAntonio Dec 4 '13 at 21:39

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.

Hint: What happens to the entries of a diagonal matrix when you square it? What is $(PDP^{-1})^2,$ expanded and simplified?

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.

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).