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

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .