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 $BA$?

  • $\begingroup$ Do $A$ and $B$ have common eigenvectors as well? $\endgroup$ – Hoda Mar 18 '14 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 '14 at 17:05

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.


If you are talking about the relationship between the eigenvalues of $AB$ and $BA$, there is a nice and well-known result, namely, $AB$ and $BA$ have identical spectra.

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 '16 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 '16 at 22:24

Actually, in your case you have specified $A^2$ which means the product of two identical matrices, and that gives us a lot of information.

So keeping it simple, I will prove that for two matrices $A, B \in \mathbb{R}^{n*n}$ and for $\vec v \in \mathbb{R}^n$ a mutual eigenvector of both with $A\vec v = \lambda v$ and $B\vec v = \mu \vec v$ for $\lambda , \mu \in \mathbb{R}$, the following is true

$AB\vec v = \lambda \mu \vec v$


$AB\vec v = A (B \vec v) = A (\mu \vec v) = \mu (A \vec v) = \mu \lambda \vec v$

So now that we have that in our hands, and since $A^2 = A*A$, (all eigenvectors of $A$ are eigenvectors of $A$), then we can show that (having $\mu_2 \in \mathbb{R}$ eigenvalue of $A^2$):

$A^2 = I \implies A^2\vec v = \vec v \implies \mu^2 = 1 \implies \mu * \mu = 1 \implies \mu = \pm1$

And to prove that there is no other values, it's clear that all vectors in $\mathbb{R}^n$ are eigenvectors to $A^2$ associated with $\lambda$, and since eigenspaces associated with different eigenvalues of the same matrix are linear independent, there is no other eienvalue than the one we calculated.


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