0

I am interested in checking the relations between the eigenvalues and eigenvectors of the two products $A B $ and $ B A$ of two $n \times n$ matrices $A$ and $B$.

First, we note that $\mbox{det}(A B) = \mbox{det}(B A)$, which implies that the product of the eigenvalues of $A B$ is equal to the product of the eigenvalues of $B A$. But we can say more about it.

Let us first consider the case when $\lambda \neq 0$ is an eigenvalue of $A B$ with the corresponding eigenvector $v$. Then by definition, $$ (A B) (v) = \lambda v, \ \ \left[ v \in \mathbf{R}^n, v \neq 0 \right] \tag{1} $$

We claim that $(\lambda, B v)$ is an eigenpair for the matrix $B A$. This can be established as follows.

$$ \lambda (B v) = B (\lambda v) = B [A B v] = (B A) (B v) $$

Thus, $$ (B A) (B v) = \lambda (B v) \tag{2} $$ and furthermore, $B v \neq 0$ because if $B v = 0$, then (1) gives $\lambda v = 0$, which cannot happen since $\lambda \neq 0$ and $v \neq 0$.

Hence, (2) establishes that $(\lambda, B v)$ is an eigenpair for $B A$.

In a very similar manner, by interchanging the roles of $A$ and $B$, we can establish the following:

If $\mu \neq 0$ is an eigenvalue for $B A$ with the eigenvector $w$, then $(\mu, A w)$ is an eigenpair for $A B$.

We consider the scenario when $\lambda = 0$ is an eigenvalue for $A B$ with the corresponding eigenvector $v$.

In this case, $\mbox{det}(A B) = 0$ which shows that $\mbox{det}(B A) = 0$.

Hence, we can conclude that $\lambda$ is also an eigenvalue of $B A$.

Does it follow that $B v$ is an eigenvector for $B A$ corresponding to $\lambda = 0$ as the eigenvalue?

We can establish that $(B A) (B v) = B [A B v] = B (0 v) = 0$ but I have a problem in showing that $B v \neq 0$. Can you throw some light on this?

If we consider a general scenario of matrices $A$ and $B$ of dimensions $(m \times n)$ and $(n \times m)$, then $A B$ will have size $(m \times m)$ and $(n \times n)$ respectively. To what extent we can extend the above calculations for the rectangular matrices $A$ and $B$? (I am yet to try calculations on these, sorry! Some insight on this is highly appreciated!)

Dr. Sundar
  • 2,677
  • It may also be noted that trace (AB) = trace(BA) – tryst with freedom Mar 31 '22 at 03:26
  • Thanks for the additional info! So the sum and product of the eigenvalues of $A B$ are the same as the sum and product of the eigenvalues of $B A$ respectively using the two equalities: (1) Trace($A B$) = Trace($B A$) and (2) Det($A B$) = Det($B A$). – Dr. Sundar Mar 31 '22 at 03:35
  • One more point is that if $A$ and $B$ share at least one eigen space, then immediately the determinant of the commutator $AB-BA$ is zero – tryst with freedom Mar 31 '22 at 03:40
  • 1
    You cannot show that $Bv$ is an eigenvector corresponding to 0. Consider the easy case where $B$ is the zero matrix. Of course, then $AB=BA$ is the zero matrix, it still shows you cannot reach the conclusion in question. – Arturo Magidin Mar 31 '22 at 04:00
  • Excellent reply! You answered my first query - thanks. – Dr. Sundar Mar 31 '22 at 04:07
  • Consider the following example: $A(x,y) =(y,0)$, $B(x,y)=(0,y)$. Then $BA(x,y).= (0,0)$, every nonzero vector is an eigenvector of zero. $AB(x,y) = (y,0)$; the eigenspace of $0$ is just ${(x,0)\mid x\in\mathbb{R}}$. So even though you have the same eigenvalues, they may not have the same geometric multiplicities. – Arturo Magidin Mar 31 '22 at 04:09
  • For square $A$ and $B$, the matrices $AB$ and $BA$ always have the same characteristic polynomial, so they have the same eigenvalues. When one of $A,B$ is nonsingular, in fact, the two product matrices are similar, so have the same Jordan form. I haven't thought about the case where both are singular. – Ted Shifrin Mar 31 '22 at 04:46
  • Following up, when both matrices are singular, the Jordan forms need not agree, so generalized eigenspaces corresponding to eigenvalue $0$ can be whatever dimensions allow. – Ted Shifrin Mar 31 '22 at 05:31
  • Related: Eigenvalues of $AB$ and $BA$ where $A$ and $B$ are rectangular matrices – Martin R Mar 31 '22 at 06:59

0 Answers0