If $n\times n$ matrix $A$ has eigenvalues $1,-1$ and $n\times n$ matrix $B$ also has eigenvalues $1,-1$, can I then say something about eigenvalues of $AB$ and $BA$?

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    $\begingroup$ Do $A$ and $B$ have common eigenvectors as well? $\endgroup$
    – Hoda
    Mar 18, 2014 at 17:00
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    $\begingroup$ Umm.. I don't know yet. My question is related to my further question math.stackexchange.com/questions/716578/… and the two matrices A and B there. I found 1,-1 be their eigenvalues. Now I have to find eigenvalues of AB and BA. I just have to take a look how to find out the eigenvectors. $\endgroup$
    – Vazrael
    Mar 18, 2014 at 17:05

2 Answers 2


If you are talking about the relationship between the eigenvalues of $AB$ and $BA$, there is a nice and well-known result, first discovered by J. J. Sylvester: $AB$ and $BA$ have identical spectra. (More generally, for any possibly rectangular matrices $A$ and $B$ with appropriate sizes, $AB$ and $BA$ share the same multi-set of nonzero eigenvalues.)

However, if you are talking about the relationship between the eigenvalues of $A,B$ and $AB$, there is only one thing that one can say:

  • the product of the eigenvalues of $AB$ is equal to the product of all eigenvalues in $A$ and $B$; this is because $\det(AB)=\det(A)\det(B)$.

Nothing further can be said without additional information. To illustrate, suppose $n=2$ and the spectra of both $A$ and $B$ are $\{1,-1\}$. The determinantal constraint in the bullet point above dictates that the spectrum of $AB$ must be $\{\lambda,\frac1{\lambda}\}$ for some nonzero $\lambda$. Now, is $\{\lambda,\frac1{\lambda}\}$ really a possible spectrum of $AB$ for every $\lambda\ne0$? The answer is yes. Let $2t=\lambda+\frac1{\lambda}$ and $$ A=\pmatrix{1&0\\ 0&-1},\ B=\pmatrix{t&1\\ 1-t^2&-t}, \ AB=\pmatrix{t&1\\ t^2-1&t}. $$ You may verify that $\operatorname{trace}(B)=0,\,\det(B)=-1,\,\operatorname{trace}(AB)=2t,\,\det(AB)=1$ and hence the characteristic polynomials of $B$ and $AB$ are respectively $x^2-1$ and $x^2-2tx+1=(x-\lambda)(x-\frac1{\lambda})$. Hence the spectrum of $B$ is indeed $\{1,-1\}$ and the spectrum of $AB$ is $\{\lambda,\frac1{\lambda}\}$. In other words, apart from the determinantal constraint mentioned in the bullet point above, the eigenvalues of $AB$ can be pretty much anything.

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    $\begingroup$ If A and B are symmetric positive definite, what one can say, for sure, about the eigenvalues of AB? $\endgroup$
    – stillanoob
    Nov 24, 2016 at 17:01
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    $\begingroup$ @ibrahim5253 When both $A$ and $B$ are positive definite, it is possible to get some bounds for the eigenvalues. E.g. using the operator norm, we get $\rho(AB)\le\|AB\|\le\|A\|\|B\|=\rho(A)\rho(B)$, where the last equality holds because $\|M\|=\rho(M)$ for any positive definite matrix $M$. $\endgroup$
    – user1551
    Nov 24, 2016 at 22:24
  • $\begingroup$ @user1551, please, what is $\rho(M)$? Could you provide a reference for these inequalities? $\endgroup$ Jan 17, 2021 at 15:04
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    $\begingroup$ @DaniloGregorinAfonso $\rho(M)$ is the standard notation for the spectral radius of a matrix $M$. It is bounded above by every submultiplicative matrix norm (including, but not limited to, operator norms). I haven't any book at hand, but I think this is mentioned (if not proved) in most textbooks on numerical linear algebra or graduate textbooks on linear algebra. I am sure that this is covered in Horn and Johnson's Matrix Analysis. $\endgroup$
    – user1551
    Jan 17, 2021 at 16:35
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    $\begingroup$ @becko It holds for all square matrices $A$ and $B$ over an algebraically field. There isn't any constraint on the eigenvalues of $A$ or $B$. More generally, if $A$ and $B$ are two rectangular matrices such that both $AB$ and $BA$ are valid matrix products, then $AB$ and $BA$ share the same multiset of nonzero eigenvalues. See this answer for a proof. $\endgroup$
    – user1551
    Nov 2, 2021 at 12:27

In general, you can almost never predict the eigenvalues of a product based on the eigenvalues of the matrices you are multiplying together. For example, consider the matrices $$A = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \qquad B = \begin{bmatrix} -1 & 1 \\ 0 & 1 \end{bmatrix} \qquad C = \begin{bmatrix} -1 & 0 \\ 1 & 1\end{bmatrix}$$ which all have eigenvalues $\pm 1$. Then $$AB = \begin{bmatrix} -1 & 1 \\ 0 & -1 \end{bmatrix}$$ has eigenvalue $-1$, $A^2 = I$ has eigenvalue $1$, and $$BC = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}$$ does not have $\pm 1$ as an eigenvalue.

The only glaring exception to this is $0$. If $0$ is an eigenvalue of either $A$ or $B$ then it is also an eigenvalue of $AB$. This is because a matrix is singular if and only if it has $0$ as an eigenvalue.


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