# Eigenvalues of $AB$ and $BA$ where $A$ and $B$ are rectangular matrices

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

Let $A$ and $B$ be $m\times n$ and $n\times m$ complex matrices, respectively, with $m < n$. If the eigenvalues of $AB$ are $\lambda_1, \ldots, \lambda_m$, what are the eigenvalues of $BA$?

If the matrices were square, then the conclusion would follow from the fact that $AB$ and $BA$ have the same characteristic polynomial. With rectangular matrices this is not going to happen; how to proceed then?

• What is your attempt? – apnorton May 24 '14 at 17:29

Let $\lambda\neq 0$ be an eigenvalue of $AB$

Then, for some non-zero $v$, $ABv=\lambda v$

Hence $BABv=\lambda Bv$

Equivalently $(BA)(Bv)=\lambda (Bv)$

Note that $Bv \neq 0$. Otherwise, $ABv=\lambda v=0$, hence $\lambda=0$

Hence $\lambda$ is a non-zero eigenvalue of $BA$

Switching $A$ and $B$ in the previous proof, it also holds that a non-zero eigenvalue of $BA$ is a non-zero eigenvalue of $AB$

Conclusion: $AB$ and $BA$ have the same non-zero eigenvalues.

• You are not assuming that $\lambda \neq 0$? – zoli May 24 '14 at 18:01
• Assumption that $\lambda \ne 0$ is crucial. Otherwise, it may happen that $Bv = 0$, which cannot be an eigenvector of $BA$. – Vedran Šego May 24 '14 at 18:11
• @VedranŠego I thought the $\lambda_i$ were non-zero. Thanks for noticing :) – Gabriel Romon May 24 '14 at 18:25

This is a totally different approach, but way more powerful.

I'm going to prove that $\chi_{BA}=(-X)^{n-m}\chi_{AB}$ by elementary means.

Let $r=\operatorname{rank}(A)$

From a well-known theorem, derive that there exists $P,Q$ invertible $m\times m$ and $n \times n$ matrices such that $$A=P\begin{bmatrix}I_r& 0\\ 0 &0\end{bmatrix}Q$$

where $I_r$ denotes the $r\times r$ identity matrix.

By changes of basis, $$B=Q^{-1}\begin{bmatrix}E& F\\ G &H\end{bmatrix}P^{-1}$$

For some submatrices $E,F,G,H$.

Note that $AB=P\begin{bmatrix}E& F\\ 0&0\end{bmatrix}P^{-1}$ and $BA=Q^{-1}\begin{bmatrix}E& 0\\ G&0\end{bmatrix}Q$.

Hence $\chi_{AB}=\det(E-XI_r)(-X)^{m-r}$ and $\chi_{BA}=\det(E-XI_r)(-X)^{n-r}$

Hence $\chi_{BA}=(-X)^{n-m}\chi_{AB}$.

The results in the two other answers are now a simple consequence of the formula.

Also, note that $BA$ will have an eigenvalue of $0$, since it's $n\times n$, but the maximum rank of each of $A$ and $b$ is $m<n$.

the characteristic polynomials of $AB$ and $BA$ are still $\lambda^{m-r}p(\lambda)$ and $\lambda^{n-r}p(\lambda), p(0) \neq 0$ the reason is $tr(AB)^k = tr(BA)^k$ for all $k$. showing the coefficients of the characteristic polynomials are the same.

• This is a cool remark, but compared to DoDom's proof above, it does not work it positive characteristic (eg over finite fields). – ACL Apr 12 '15 at 16:33

Eigenvalues are roots of characteristic polynomial. We want to find the connction between characteristic polynomials of AB and BA. Let $$\chi_M(x)$$ denotes a characteristic polynomial $$\chi_M(x) = det(x - M)$$

Lets prove the fact: For square matrices $$A$$ and $$B$$ holds $$det(AB - x) = det(BA - x) \Leftrightarrow \chi_{AB}(x) = \chi_{BA}(x)$$.

If $$det(A) \neq 0$$ then it follows from $$det(AB - x) = det(A^{-1}A)det(AB - x) = det(A^{-1})det(AB - x)det(A) = det(BA - x)$$.

If $$det(A) = 0$$ there are finite number of such $$s \in \mathbb R$$ that $$\chi_A(s)=0$$ because $$\chi_A(s)$$ is a finite-degree polynomial. Then there are infinite number of such $$s$$ that $$\chi_A(s) \neq 0$$. For all such $$s$$ we know $$\chi_{(A-s)B}(x) = \chi_{B(A-s)}(x)$$ as a result of a previous case. For every fixed $$x$$ we see two finite-degree polynomials ($$x$$ is fixed, $$s$$ is variable) $$\chi_{(A-s)B}(x)$$ and $$\chi_{B(A-s)}(x)$$ which are equal in infinite number of points. Then we conclude they are equal at every $$s$$. At $$s = 0$$ we get the result $$\chi_{AB}(x) = \chi_{BA}(x)$$ at every $$x$$.

For square matrices we are done!

Key fact (proof below): If $$A$$ is $$m\times n$$, $$B$$ is $$n\times m$$ and $$n \geq m$$ then $$\chi_{BA}(x) = \lambda^{n-m}\chi_{AB}(x)$$.

Consider $$n\times n$$ matrices $$A' = \left(\dfrac{A}{0}\right)$$ and $$B' = (B\mid0)$$. We just put zero rows and columns to make matrices $$n\times n$$.

First, $$B'A' = BA \Rightarrow x - B'A' = x - BA \Rightarrow \chi_{B'A'}(x) = \chi_{BA}(x)$$

Second, $$A'$$ and $$B'$$ are square matrices. Then due to the fact above we have $$\chi_{B'A'}(x) = \chi_{A'B'}(x)$$.

Third, $$\chi_{A'B'}(x) = det(x - A'B') = det\begin{pmatrix}x - AB & 0 \\ 0 & \begin{matrix}x & 0 & \ldots & 0 \\ 0 & x & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & x \\\end{matrix}\end{pmatrix} = det(x - AB)x^{n - m} = x^{n-m}\chi_{AB}(x)$$

So, we see $$\chi_{BA}(x) = \chi_{B'A'}(x) = \chi_{A'B'}(x) = x^{n-m}\chi_{AB}(x)$$. Then wee see all eigenvalues of $$AB$$ are eigenvalues of $$BA$$ and other $$n-m$$ eigenvalues are zeros in $$BA$$.