5
$\begingroup$

The matrix $A \in M_3(\mathbb{R})$ satisfy $A^t=A$ and $(1,2,1), (-1,1,0)$ are eigenvectors of $A$. Which vector is also an eigenvector of $A$? Alternatives: $(0,0,1)$; $(1,1,-3)$; $(1,1,3)$; There is no other eigenvector.

The problem with this exercise is that I don't know the matrix $A$, and I don't have any eigenvalue to start with. I can get a matrix with less variables using $A = A^t$, but there's still 6 variables. Any tips or guidance is appreciated.

$\endgroup$
  • $\begingroup$ There is some sloppiness in the language here. Of course there are more eigenvectors. In fact, there is an infinite amount of them. For instance, for every nonzero $\lambda$, $\lambda (1, 2, 1)$ is an eigenvector. $\endgroup$ – Andreas Rejbrand Feb 17 at 12:16
  • 1
    $\begingroup$ The question only makes sense if the eigenvalues are distinct. Counter-example: if $A = 0$ every vector is an eigenvector. $\endgroup$ – alephzero Feb 17 at 13:49
6
$\begingroup$

Since $A$ is symmetric, the eigenvectors (for distinct eigenvalues) are orthogonal.

So, find which of the vectors is orthogonal to the first two.

(1,1,-3) is.

$\endgroup$
  • $\begingroup$ I think I need to rethink it. I'm leaning towards $(1,1,-3)$ because it's orthogonal to both. Will edit. $\endgroup$ – Chris Custer Feb 17 at 7:01
  • 3
    $\begingroup$ The other two could be eigenvectors. If so, the matrix would have to be a multiple of the identity, but I don’t see anything in the problem statement that would preclude this. $\endgroup$ – amd Feb 17 at 7:08
  • $\begingroup$ @amd true. Good catch. $\endgroup$ – Chris Custer Feb 17 at 7:14
  • 1
    $\begingroup$ @amd the orthogonal one must be an eigenvector though, if I'm not mistaken. $\endgroup$ – Chris Custer Feb 17 at 7:23
  • 1
    $\begingroup$ @ChrisCuster Yeah the problem could probably have been better stated as "which vector is also necessarily an eigenvector" $\endgroup$ – angryavian Feb 17 at 7:45
5
$\begingroup$

Hint: the condition $A^t = A$ allows you to use the spectral theorem.

Hint: Specifically, the spectral theorem implies there is an orthonormal basis of eigenvectors of $A$.

$\endgroup$
  • $\begingroup$ That is correct, but unless the eigenvalues are distinct the theorem does not say that no other vectors except for the basis vectors are eigenvectors. $\endgroup$ – alephzero Feb 17 at 15:21
  • $\begingroup$ @PaulSinclair No, I think "no other eigenvector" is a possible "choice" for this multiple choice question. $\endgroup$ – angryavian Feb 17 at 17:29
  • $\begingroup$ @angryavian - you are correct. As Chris Custer pointed out below his answer, the two eigenvectors given are not orthogonal. Therefore their eigenvalues have to be the same. I mis-interpreted the problem. $\endgroup$ – Paul Sinclair Feb 17 at 17:37

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.