# Do the non-zero eigenvalues of AB and BA have the same algebraic multiplicity (for AB and BA not square)?

I know that if A and B are square nxn matrices, then AB and BA have the same characteristic polynomial and thus the same eigenvalues (and same algebraïc multiplicity).

I'm wondering though if this can be generalized: if A is a nxm matrix and B a mxn matrix, then AB is a nxn matrix and BA a mxm matrix. So my question is: will the eigenvalues of AB and BA, that differ from zero, have the same algebraïc multiplicity?

• How on earth can you compare two polynomials while "ignoring the exponent of x"? Commented Mar 17, 2013 at 10:56
• @ChrisEagle He clarified at the end what he actually meant. Commented Mar 17, 2013 at 10:57
• I guess this theorem will be helpful. Commented Mar 17, 2013 at 11:04
• Yes, I messed a little up in formulating the question. I edited my post now ;). Commented Mar 17, 2013 at 11:04
– Did
Commented Mar 17, 2013 at 11:05

Suppose $A$ has $d$ more rows than columns, and therefore that $B$ has $d$ more columns than rows. Add $d$ zero columns to $A$, and $d$ zero rows to $B$, to get square matrices $A',B'$. The product $A'B'$ is identical to $AB$, while $B'A'$ is obtained from $BA$ by adding $d$ zero rows and $d$ zero columns. Since $B'A'$ is block diagonal (actually block-triangular would have sufficed), the characterisitic polynomial of $B'A'$, which is equal to that of $A'B'$ by the result for square matrices, is $X^d$ times the characteristic polynomial of $BA$. Therefore what you guessed is indeed true: $$\chi_{BA}=X^d\chi_{AB}.$$
Suppose $$A$$ is an $$n\times n$$ matrix with the black decomposition \begin{align} A=\begin{pmatrix} A_{11} & A_{12}\\ A_{21} & A_{22} \end{pmatrix} \end{align} where $$A_{ii}$$ is a squared matrix of size $$n_i\times n_i$$, $$i = 1,2$$.
If $$\operatorname{det}(A_{11})\neq0$$ then \begin{align} \operatorname{det}(A)=\operatorname{det}(A_{11})\cdot\operatorname{det}(A_{22}-A_{21}A^{-1}_{11}A_{12})\tag{0}\label{zero} \end{align} If $$\operatorname{det}(A_{22})\neq0$$ then \begin{align} \operatorname{det}(A)=\operatorname{det}(A_{22})\cdot\operatorname{det}(A_{11}-A_{12}A^{-1}_{22}A_{21})\tag{0'}\label{zerop} \end{align}
Suppose $$A$$ and $$B$$ are matrices of size $$k\times \ell$$ and $$\ell\times k$$ respectively. Consider the matrix $$L$$ of size $$n\times n$$, where $$n=k+\ell$$, given by \begin{align} L:=\begin{pmatrix} I_k & A\\ B & I_\ell\end{pmatrix} \end{align} where $$I_k$$ and $$I_\ell$$ are the identity matrices of size $$k$$ and $$\ell$$ respectively. Then, by \eqref{zero} and \eqref{zerop} $$\operatorname{det}(L)=\operatorname{det}(I_k - AB)=\operatorname{det}(I_\ell - BA)\tag{1}\label{one}$$
Let $$p_{AB}$$ and $$p_{BA}$$ denote the characteristic polynomials of squared matrices $$AB$$ and $$BA$$ respectively. Then, by \eqref{one}, for any $$t\neq0$$ \begin{align} p_{AB}(t)&=(-1)^k\operatorname{det}(tI_k - AB)=(-t)^k\operatorname{det}(I_k -\tfrac1tAB)\\ &=(-t)^k\operatorname{det}(I_\ell - \tfrac1t BA)=(-1)^k t^{k-\ell}\operatorname{det}(tI_\ell - BA)\\ &= (-t)^{k-\ell}p_{BA}(t) \end{align} From this identity, it follows that or any $$t_*\neq0$$, $$t_*$$ is an eigenvalue of $$AB$$ iff $$t_*$$ is an eigenvalue of $$BA,$$ and moreover, the (algebraic) multiplicity of $$t_*$$ as an eigenvalue of $$AB$$ is the same as an eigenvalue of $$BA$$.