8
$\begingroup$

Is there a relation between the eigenvectors of matrix $A$ and of matrix $A^TA$?

This question is related to eigenvectors, not eigenvalues. Further, the sizes of matrices $A$ and $A^TA$ are different if $A$ is not a square matrix. Thus, my question has no relation to the question about eigenvalues of $A$ ans $A^T$.

$\endgroup$
3
  • $\begingroup$ Do you mean eigenvalues are same? $\endgroup$
    – Creator
    May 23, 2015 at 5:28
  • $\begingroup$ If you're asking if $A$ and $A^TA$ share the same set of eigenspaces, then this really comes down to "do $A$ and $A^T$ share the same set of eigenspaces?" And the answer to that is "no". Just consider $\pmatrix{1 & 1 \\ 0 & 0}$ and $\pmatrix{1 & 0 \\ 1 & 0}$. The eigenvalues on the other hand... $\endgroup$
    – user137731
    May 23, 2015 at 5:34
  • $\begingroup$ $Ax = \lambda_1 x, A^TAy = \lambda_2 y$ $\to A^T \lambda_1 y = \lambda_2 y$ I think? $\endgroup$
    – BCLC
    May 23, 2015 at 5:47

2 Answers 2

4
$\begingroup$

Let $A=U\Lambda V^\top$, $U$, $\Lambda$ and $V$ are respectively left eigenvectors, eigenvalues and right eigenvectors.

$$A^\top A=U\Lambda V^\top V \Lambda U^\top=U\Lambda^2 U^\top$$

So what are the eigenvalues of $A^\top A$ ?

$\endgroup$
3
  • $\begingroup$ You mean that $A=U\Lambda V^T$ is the SVD of $A$, not its spectral decomposition. $\endgroup$ May 23, 2015 at 14:30
  • $\begingroup$ Yes, indeed. It's the svd $\endgroup$
    – davcha
    May 23, 2015 at 14:31
  • $\begingroup$ Ok, so the columns of $U$ and $V$ are left and right singular vectors of $A$, not eigenvectors. $\endgroup$ May 23, 2015 at 14:44
3
$\begingroup$

There is no simple relation between eigenvectors of $A$ and of $A^*A$ unless $A$ is normal. For example, if $$ A=\pmatrix{0&1\\0&0}, $$ $A$ has only one dimensional eigenvector subspace spanned by $[1,0]^T$, but $A^*A=0$ so any nonzero 2-vector is its eigenvector. Even if $A$ is diagonalizable, the eigenspaces of $A$ and $A^*A$ can be completely different.

If $A$ is normal, then $A=UDU^*$ for a unitary $U$ and diagonal $D$, so $A^*A=U|D|^2U^*$ is the spectral decomposition of $A^*A$.

However, there is a simple relation between left singular vectors of $A$ and the eigenvectors of $A^*A$ as indicated in another answer.

$\endgroup$
2
  • $\begingroup$ The diagonal matrix gets squared. $\endgroup$
    – Ian
    May 23, 2015 at 14:46
  • $\begingroup$ @Ian Yep thanks! $\endgroup$ May 23, 2015 at 14:48

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.