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Let $A$ be a real $m\times n$ matrix, $m\leq n$. Now, what can we say about the eigenvectors of $A^TA$ and $AA^T$?
First of all, the rank of $A\leq \min(m,n)=m$.
Now, $R=A^TA$ is $n\times n$ matrix and $L=AA^T$ is $m\times m$ matrix.
What can we say about their eigenvectors and eigenvalues? (Like, how many eigenvectors corresponding to $0$ eigenvalue)

I know that the eigenvector of L corresponding to a non-zero eigenvalue is an eigenvector of R with the same eigenvalue and vice-versa. So if one has exactly $x$ eigenvectors corresponding to non-zero eigenvalues, then the other one also has exactly $x$ eigenvectors corresponding to those non-zero eigenvalues. But what about the rest (if any)?
Note: Sorry for the confusion, I meant equivalence only. For every eigenvector (non-zero eigenval.) of $L/R$, there is a corresponding eigenvector for the other.
Please give the reasoning behind the claims too, thank you!

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    $\begingroup$ No, the eigenvectors of $L$ for nonzero eigenvalues are not eigenvectors of $R$: they don't even have the same dimension if $m \ne n$. $\endgroup$ Nov 5 '20 at 18:19
  • $\begingroup$ Echoing @RobertIsrael. They're directly related, but not equivalent. $\endgroup$ Nov 5 '20 at 18:20
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What is true is that if $x$ is an eigenvector of $AA^T$ for nonzero eigenvalue $\lambda$, then $A^T x$ is an eigenvector of $A^T A$ for $\lambda$, and if $y$ is an eigenvector of $A^T A$ for nonzero eigenvalue $\lambda$, then $A y$ is an eigenvector of $A A^T$ for $\lambda$.

There are $r$ linearly independent eigenvectors $u_n$ of $A A^T$ for nonzero eigenvalues, where $r$ is the rank of $A$, and $r$ linearly independent eigenvectors $v_n$ of $A^T A$ for nonzero eigenvalues. These may be chosen so that $u_n = A v_n$ for each $n$, and the corresponding eigenvalues are the same. Another $n-r$ linearly independent eigenvectors of $A^T A$ are for eigenvalue $0$, and $m-r$ linearly independent eigenvectors of $A A^T$ are for eigenvalue $0$. These form bases of the null spaces of $A$ and $A^T$ respectively.

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  • $\begingroup$ Thanks! Can you please give an explanation/reasoning behind the claims? $\endgroup$ Nov 5 '20 at 18:34
  • $\begingroup$ Namely, why are rank of $A$ and number of eigenvectors of $L$, related. And how can we say that there are $n-r$ and $m-r$ linearly independent eigenvectors for those. $\endgroup$ Nov 5 '20 at 18:37

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