1
$\begingroup$

Let $M$ be a $100\times 2$ matrix. Is it true that $MM^{T}$, which is $100\times 100$ has the same number of eigenvectors with non-zero eigenvalues as $M^T M$ (at most $2$ eigenvectors)? How many eigenvectors in total does $MM^{T}$ have? If I'd have to guess, it'd be $100$...

$\endgroup$
  • $\begingroup$ The matrices $MM^T\in\mathbb{R}^{100\times 100}$, $M^TM\in \mathbb{R}^2$ are both symmetric and positive definite, i.e. their eigenvalues are $\lambda\geq 0$. The first one has 100 eigenvalues, and the second 2 eigenvalues. $MM^T$ and $M^TM$ share the same nonzero eigenvalues. $\endgroup$ – xecafe Mar 24 '15 at 19:31
1
$\begingroup$

The null space of $MM^T$ have atleast $98$ linearly independent eigenvectors. The total number of linearly independent eigenvectors is $98$ + number of linearly independent eigenvectors of the matrix $M^TM$.

Otherwise, suppose $x$ is an eigenvector of $M^TM$ corresponding to the nonzero eigenvalue $\lambda$, then $Mx$ an eigenvector of $MM^T$ corresponding to the same eigenvalue. For, if $M^TMx = \lambda x$, then $MM^TMx = \lambda Mx$. Since $x$ is an element of range of $M^T$, $Mx\neq 0$.

If $M^TM$ has two distinct eigenvalues or $M^TM = 0$, then it is clear.

Also, if $x$ and $y$ are linearly independent eigenvectors of $M^TM$ corresponding to the eigenvalue $\lambda$, then $Mx$ and $My$ are linearly independent eigenvectors of $MM^T$ corresponding to $\lambda$. For, if $Mx= \mu My$, then $M^T(Mx - \mu My) = 0$. This implies that $x$ an $y$ are linearly dependent. Similarly we can prove the other way. i.e. if $x$ and $y$ are linearly independent eigenvectors of $MM^T$, then $M^Tx$ and $M^Ty$ are linearly independent eigenvector of $M^TM$ with the same eigenvalue.

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

When $A$ is $m \times n$ and $B$ is $n \times m$, it's always true that $AB$ and $BA$ have the same nonzero eigenvalues, with the same algebraic and geometric multiplicities. This has been discussed here a number of times, e.g. here and here.

| 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.