Eigen values of AB and BA

Question

let A be a linear transformation from $R^n$ to $R^m$, and B be a linear transformation from $R^m$ to $R^n$, it's easy to show that AB and BA has same eigen-value(except $0$).
But my question is how to show that the multiplicity of eigen-values are the same? can anyone give a proof just from the theory of linear transformation? I mean, without matrix computation.

Answer 1

Just in case you want the result for geometric multiplicity (dimension of eigenspaces) instead of (or in addition to) algebraic multiplicity, here's a matrix-free proof of that result.

Consider any non-zero eigenvalue $\lambda$ of $AB$ and the corresponding eigenspace $E$. Also, let $F$ be the eigenspace of $BA$ for the same eigenvalue $\lambda$. (If you didn't already know that $\lambda$ is an eigenvalue of $BA$, then I'd allow the possibility that $F=\{0\}$; that wouldn't damage the following argument.) For any $x\in E$, we have $ABx=\lambda x$, and therefore, applying $B$ to both sides, we also have $BABx=\lambda Bx$. This says that $Bx\in F$. So we've shown that $B$ maps $E$ into $F$. Furthermore, this map is injective; non-zero vectors $x\in E$ have non-zero images $Bx\in F$, because $ABx=\lambda x\neq0$. (This is where we use that $\lambda\neq0$.) Since we have an injective linear map from $E$ into $F$, it follows that $\dim(E)\leq\dim(F)$. A symmetrical argument, interchanging the roles of $A$ and $B$) shows that $\dim(F)\leq\dim(E)$. Therefore the two dimensions are equal.

Answer 2

If $n=m$: We want to show that characteristic polynomial of $AB=$ the characteristic polynomial of $BA$, i.e $det(AB-xI_n)=\det(BA-xI_n)$

• First case : if $A$ is invertible: $$\det(AB-xI_n)=\det(A(B-xA^{-1}))=\det(A)\det(B-xA^{-1})=\det(B-xA^{-1})\det(A)=\det((B-xA^{-1})A)=\det(BA-xI_n)$$
• Second case: if $A$ is invertible:
  Let $\{0,\lambda_2,\lambda_3,\ldots,\lambda_r\}$ the eigenvalues of $A$, with $0<|\lambda_1|<\ldots <|\lambda_r|$, now remarking that if$ \alpha\in \Bbb R$ such that $0<\alpha<|\lambda_1|$ the $x$ is not an eigenvalue of $A$ and $A-\alpha I_n$ is invertible, it is easy to show that there is $N\in \Bbb N$ such that for all $p\leq N$, $0<\dfrac{1}{p}<|\lambda_1|$, then for all $p\geq N$, the matrix $A_p=A-\dfrac{1}{p}I_n$ is invertible. By the first case for all $p\geq N$, $\det(A_pB-xI_n)=\det(BA_p-I_n)$ passing to the limit as $p\to \infty$ (and using the fact that $\det$ is continuous and $A_p\to A$ ) we obtain $\det(AB-xI_n)=\det(BA-xI_n)$.

Answer 3

Here is a way that uses matrix computations (that works for all $n,m$):

Let $R_1 = \begin{bmatrix} I & 0 \\ -A & I \end{bmatrix} $, $R_2 = \begin{bmatrix} I & -B \\ 0 & I \end{bmatrix}$, $M=\begin{bmatrix} I & B \\ A & I \end{bmatrix}$.

Note that $R_1 M = \begin{bmatrix} I & B \\ 0 & I-AB \end{bmatrix}$, $R_2 M = \begin{bmatrix} I-BA & 0 \\ A & I \end{bmatrix}$, and $\det R_k = 1$.

Consequently, we have $\det (I-AB) = \det(I-BA)$ for all appropriately sized $A,B$.

Now suppose $\lambda \neq 0$, and replace $A$ by ${1 \over \lambda } A$, which gives $\det (I-{1 \over \lambda }AB) = \det(I-{1 \over \lambda }BA)$, or ${1 \over \lambda^m } \det (\lambda I-AB) = {1 \over \lambda^n } \det (\lambda I-BA) $.

Note that $p_{AB}(\lambda)= \det (\lambda I-AB) $ and $p_{BA}(\lambda)= \det (\lambda I-BA) $ are both polynomials, hence it follows that $(\lambda-\mu)^k$ divides $p_{AB}$ iff it divides $P_{BA}$, and so the non-zero eigenvalues of $AB$ and $BA$ have the same algebraic multiplicity.

(Note: for a truly 'matrix computation-free' proof, one could show that $\dim \ker (\lambda I-AB)^k = \dim \ker (\lambda I-BA)^k$ for all $\lambda \neq 0$, and $k$, but this is far more tedious.)