# If $A,B$ have eigenvalues $\pm1$, what can I say about the eigenvalues of $AB,BA$?

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 $BA$?

• Do $A$ and $B$ have common eigenvectors as well?
– Hoda
Mar 18, 2014 at 17:00
• 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. Mar 18, 2014 at 17:05

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.

• If A and B are symmetric positive definite, what one can say, for sure, about the eigenvalues of AB? Nov 24, 2016 at 17:01
• @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 $M$. Nov 24, 2016 at 22:24
• @user1551, please, what is $\rho(M)$? Could you provide a reference for these inequalities? Jan 17, 2021 at 15:04
• @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. Jan 17, 2021 at 16:35
• @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. 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.