This question is a generalisation of Eigenvalues of $AB$ and $BA$ where $A$ and $B$ are rectangular matrices which itself is a generalisation of Eigenvalues of $AB$ and $BA$ where $A$ and $B$ are square matrices.

Let $A$ be an $m \times n$ matrix and $B$ an $n \times m$ matrix. Obviously, the matrix products $AB$ and $BA$ are possible. Assume $n \leq m$, such that $AB$ is a weakly larger matrix than $BA$.


  1. The rank of both $AB$ and $BA$ is at most $n$ (link 1)
  2. The number of non-zero eigenvalues of both $AB$ and $BA$ is at most $n$ (link 2)
  3. If the eigenvalues of $AB$ are $\lambda_1, \ldots, \lambda_n$, the eigenvalues of $BA$ are also $\lambda_1, \ldots, \lambda_n$ (link 3).


  1. If the singular values of $AB$ are $\sigma_1, \ldots, \sigma_n$, what can be said about the singular values of $BA$?
  2. What does Fact 3, compared with the answer to Question 1, say about the differences and the similarities between eigenvalues and singular values?
  • $\begingroup$ @Ian Are the eigenvalues of $(AB)(B^TA^T)$ related to those of $(A^TB^T)(BA)$? I'm not seeing the connection. $\endgroup$ – Erick Wong Sep 28 '17 at 8:08
  • $\begingroup$ @ErickWong You're right, I made a simple mistake. $\endgroup$ – Ian Sep 28 '17 at 15:42
  • $\begingroup$ There is a relation between the singular values but a subtle one: Singular values of both $AB$ and $BA$ satisfy the same set of linear inequalities, called Horn inequalities, in terms of singular values of $A$ and $B$. I can tell you more if you are interested. $\endgroup$ – Moishe Kohan Oct 1 '17 at 21:55

There is almost no relationship. For example, if we take $$A = \begin{bmatrix}x & 1 \\ 0 & 0\end{bmatrix}, \quad B = \begin{bmatrix}0 & 0 \\ 1 & y \end{bmatrix}$$ then the singular values of $AB$ are the square roots of the eigenvalues of $$(AB)^{\mathsf T} AB = \begin{bmatrix}1 & y \\ y & y^2\end{bmatrix}$$ so they are $\sqrt{1+y^2}$ and $0$. Similarly, the singular values of $BA$ are $\sqrt{1+x^2}$ and $0$. Even in this simple example, the nonzero singular value in one case can vary pretty much independently of the other case. (They must both be at least $1$, but we can tweak that by changing the $1$ in the matrices to some small $\epsilon>0$.)

By taking determinants, we can conclude that the product of the singular values of $AB$ is $\det(AB)$, while the product of the singular values of $BA$ is $\det(BA)$. So if $A$ and $B$ are both square matrices, the singular matrices in both cases have an equal product $\det(A)\det(B)$, which is some amount of dependence.

On the other hand, by taking tensor products of the construction above, we can start with $2n \times 2n$ square matrices $A$ and $B$ where

  • $AB$ has singular values $\sigma_1, \sigma_2, \dots, \sigma_n, 0, 0, \dots, 0$,
  • $BA$ has singular values $\sigma'_1, \sigma'_2, \dots, \sigma'_n, 0, 0, \dots, 0$,
  • and these are free to vary independently of each other.
  • $\begingroup$ Thank you. This is somewhat surprising to me as I previously thought of singular values as just a generalisation of eigenvalues. You've shown that, although $AB$ and $BA$ have identical eigenvalues, their singular values are independent. Is there an intuitive/visual way to understand the difference between singular values and eigenvalues? $\endgroup$ – LBogaardt Oct 1 '17 at 8:24
  • 2
    $\begingroup$ The eigenvalue decomposition involves finding a single basis (of any kind) on which the matrix plays nicely. The singular value decomposition involves finding two different orthonormal bases on which the matrix plays nicely. In the end, it turns out that $A$ maps the eigenbasis of $BA$ to multiples of the eigenbasis of $AB$, hence your fact 3. For the equivalent to hold for SVD, $A$ would have to map orthonormal vectors to other orthonormal vectors, which would require special properties of $A$. $\endgroup$ – Misha Lavrov Oct 1 '17 at 16:31

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