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!

  • 4
    $\begingroup$ 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

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.


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

  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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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