1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – DonAntonio Dec 4 '13 at 21:39
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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).

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.