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I want to prove that

$$\det \left( x I_m - A B \right) = x^{m-n} \det \left( x I_n - B A \right)$$

where $A \in \mathbb{F}^{m \times n}$ and $B \in \mathbb{F}^{n\times m}$.

It is easy to show that $0$ has algebraic multiplicity of at least $m-n$ for $AB$, but how can I show that the other eigenvalues of $AB$ are actually eigenvalues of $BA$ ? I know that the sum of eigenvalues of $AB$ and $BA$ are the same, because $\operatorname{trace}(AB)=\operatorname{trace}(BA)$, but ...

I appreciate any help, thanks ! ${{{}}}$

Rodrigo de Azevedo
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the8thone
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    I think you want $AB$ in one determinant and $BA$ in the other. – Pedro Jun 23 '14 at 03:18
  • okay thanks ! I just corrected it – the8thone Jun 23 '14 at 03:19
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    @the8thone I think my proof here: http://math.stackexchange.com/a/822014/152030 works even when $A$ and $B$ are not of the same order (but compatible for multiplication). – M. Vinay Jun 23 '14 at 03:26
  • Thanks ! but what if we have $m=5$ & $n=4$ & $\chi_{AB}=x^2(x-1)(x-2)^2$ and $\chi_{BA}=x(x-1)^2(x-2)$... I know that this can not happen, but why ? – the8thone Jun 23 '14 at 03:33
  • This is a reformulation of Sylvester's theorem. You find a proof (here)[http://en.wikipedia.org/wiki/Sylvester%27s_determinant_theorem ]. – Fabian Jun 23 '14 at 03:35

2 Answers2

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

Gabriel Romon
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This proof is nothing new. It is a reformulation of Sylvester's determinant theorem as suggested in Fabian's comment.

Assume $x \ne 0$, we have

$$ \begin{bmatrix}I_n & B\\ A & xI_m\end{bmatrix} \begin{bmatrix}I_n & -B\\ 0 & I_m\end{bmatrix} = \begin{bmatrix}I_n & 0\\ A & xI_m - AB\end{bmatrix}$$ This implies $$\det(xI_m - AB) = \begin{vmatrix}I_n & 0\\ A & xI_m - AB\end{vmatrix} =\begin{vmatrix}I_n & B\\ A & xI_m\end{vmatrix}\tag{*1} $$ Notice $$\begin{vmatrix}I_n & B\\ A & xI_m\end{vmatrix} = (-)^{mn} \begin{vmatrix}A & xI_m\\I_n & B\end{vmatrix} = \begin{vmatrix}xI_m & A\\B & I_n\end{vmatrix} = x^{m+n}\begin{vmatrix}I_m & x^{-1}A\\x^{-1}B & x^{-1}I_n\end{vmatrix} $$ Apply $(*1)$ with $x^{-1}B, x^{-1}A, x^{-1}I_n$ taking the roles of $A, B, xI_m$, we get

$$\begin{align}\det(xI_m-AB) &= x^{m+n}\det(x^{-1}I_n - x^{-2}BA) = x^{(m+n)-2n}\det(x I_n - BA)\\ &= x^{m-n}\det(xI_n - BA)\end{align}$$

This leave us with the special case $x = 0$,

  • if $m = n$, the relation at hand reduces to $\det(-AB) = \det(-BA)$ which is trivially true.
  • if $m > n$, LHS reduces to $\det(-AB)$ where $AB$ is a $m \times m$ matrix. Since the rank of $A$ and $B$ is at most $n < m$, this determinant is zero. RHS is zero because of the $x^{m-n}$ factor.

Combine all these, we have

$$\det(x I_m - AB ) = x^{m-n} \det(x I_n - BA)$$

unless $x = 0$ and $m < n$ where the RHS is an undefined expression.

achille hui
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