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Let $A$ be a real $n\times n$ matrix. How are the eigenvalues of $A$ and $AA^T$ related?

What I have come up with so far is that if we let $\lambda_1,\ldots,\lambda_n$ denote the eigenvalues of $A$, then since $$ \det(AA^T) = \det(A)^2 $$ we can deduce the product of the eigenvalues of $AA^T$ is the product of the eigenvalues of $A$ squared.

This lead me to believe that $\lambda_1^2,\ldots,\lambda_n^2$ are the eigenvalues of $AA^T$, but I have not been able to prove it.

Any help is appreciated!

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    $\begingroup$ This is false. $AA^T$ always has real eigenvalues, while the eigenvalue of $A$ might not be real nor pure imaginary. $\endgroup$ – JWL Feb 19 '17 at 11:46
  • $\begingroup$ this is not true, a sufficient condition for it to be true is that $A$ is symmetric, but I'm not sure if that's necessary or not. Can you come up with a counter example? $\endgroup$ – Thoth Feb 19 '17 at 11:46
  • $\begingroup$ It is true that $\lambda_i^2$ are the eigenvalues of $A^2$. $\endgroup$ – JWL Feb 19 '17 at 11:47
  • $\begingroup$ @JWL Right, that makes sense. So there is not really a way of finding the eigenvalues of $A$, when I only know the eigenvalues of $AA^T$, right? $\endgroup$ – user114158 Feb 19 '17 at 11:53
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The eigenvalues of $A A^T$ are nonnegative real numbers. Their square roots are called the singular values of $A$ (and $A^T$). Like the eigenvalues of $A$, they multiply out to the determinant. Nonetheless your statement usually fails. A rotation by $\pi/2$ in $\mathbb{R}^2$ shows the point easily enough: the eigenvalues are $\pm i$, their square is $-1$, but $A A^T=I$, whose eigenvalues are $1$. Requiring the eigenvalues to be real doesn't fix the matter, either.

You can get a relationship when $A$ is normal: in this case $A$ and $A^*$ (the conjugate transpose) commute, so they share eigenvectors. The eigenvalues of $A^*$ are the conjugates of the eigenvalues of $A$, however, even when $A$ was real to begin with. Thus for an eigenvector $x$ of $A^*$ with eigenvalue $\lambda$ you have $A A^* x = \lambda A x = \lambda \overline{\lambda} x = |\lambda|^2 x$. But even now taking the square root removes whatever complex phase that $\lambda$ may have had.

You could have proved the same thing from the assumption of unitary diagonalizability...but in fact the (generalized) spectral theorem tells us that this is equivalent to normality.

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Recall that $A$ and $A^T$ have the same set of eigenvalues.

Since, for $\lambda \in \mathbb{R}$ we have that $Ax = \lambda x$ and $A^Tx = \lambda x$ we obtain

$$A^TAx = A^T(Ax) = A^T\lambda x = \lambda(A^Tx) = \lambda^2x$$

and similarly $$AA^Tx = A(A^Tx) = A\lambda x = \lambda(Ax) = \lambda^2x$$

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    $\begingroup$ Why do they share eigenvectors? (Answer: they don't.) $\endgroup$ – Ian Feb 19 '17 at 11:54
  • $\begingroup$ $A A^T$ is symmetric, so all of its eigenvalues are real. What if $\lambda^2$ is not real? Then it cannot be an eigenvalue of $A A^T$. $\endgroup$ – littleO Feb 19 '17 at 11:58
  • $\begingroup$ @Ian does it mean that in this case we would need $A$ to be normal for $A$ and $A^T$ to share the eigenvectors? $\endgroup$ – Maciej Caputa Feb 19 '17 at 12:06
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    $\begingroup$ If $A$ is normal then they share eigenvectors, but then the eigenvalues are conjugates, so that the corresponding eigenvalue of $A A^T$ is $|\lambda|^2$. $\endgroup$ – Ian Feb 19 '17 at 12:17

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