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 $100$...
Asked
Active
Viewed 402 times
1
-
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. – petem Mar 24 '15 at 19:31
2 Answers
1
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.
mrka
- 648
0
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.
Robert Israel
- 448,999