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There is a problem in Hoffmann Kunze:

Show that $AB$ and $BA$ have the same characteristic polynomials,where $A,B$ are both $n\times n$ matrices in $\mathbb F$.

If $A$ or $B$ is invertible it can be easy done:

$det(AB-xI).det(A)=det(ABA-xA)=det(A).det(BA-xI)$ and if $detA\neq 0$,we can cancel it to get our desired result but what about the case when both $detA $ and $detB$ are $0$.How to proceed in that case?

By the way the case $n=2$ is easy as $p_A(x)=x^2-tr(A)x+det(A)$ and so characteristic polynomial of $AB$ and $BA$ must be same as they have the same trace and same determinant.But this idea cannot be applied for higher order matrices.

Kishalay Sarkar
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  • If you know about exterior algebra, you can extend that $n=2$ argument to any $n$. – Angina Seng May 07 '20 at 15:22
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    View the coefficients of $A$ and $B$ as variables, say $A = (a_{ij})$ and $B = (b_{ij})$. You then have $\det(AB - xI) = \det(BA - xI)$ as an identity over the field $F = \Bbb F(a_{11}, \dotsc, a_{nn}, b_{11}, \dotsc, b_{nn}, x)$, by what you already know: because $A$ is invertible over $F$. – WhatsUp May 07 '20 at 15:22
  • See this or https://math.stackexchange.com/questions/17831/sylvesters-determinant-identity – cqfd May 07 '20 at 15:46

2 Answers2

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A bit tricky way, but which it even can be applied when $A$ and $B$ are not square matrices, so $A$ is $n\times m$ and $B$ is $m\times n$, is the following:

Consider the identity of the product of matrices with dimensions $(n+m)\times (n+m)$: $$\begin{pmatrix} I_n & -A \\ B & I_m \end{pmatrix}= \begin{pmatrix} I_n & 0 \\ B & I_m \end{pmatrix} \begin{pmatrix} I_n & -A \\ 0 & I_m+BA \end{pmatrix}= \begin{pmatrix} I_n+AB & -A \\ 0 & I_m \end{pmatrix}\begin{pmatrix} I_n & 0 \\ B & I_m \end{pmatrix}.$$

Now, using the following result on the determinant of "block" triangular matrices $$\text{det}\begin{pmatrix} A & 0 \\ B & C\end{pmatrix}=\text{det}(A)\text{det}(C)$$ we get that $$\text{det}(I_n+AB)=\text{det}(I_m+BA).$$

Finally, one gets from this that $$(-x)^{m}\text{det}(AB-xI_n)=(-x)^{n} \text{det}(BA-xI_m).$$ which, in your case $n=m$, gives the result.

Nulhomologous
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How about this? introduce a new indeterminate $t$, and consider $(A+tI)B-xI$ and $B(A+tI)-xI$. Over the extension field $\Bbb F(t)$ the matrix $(A+tI)$ is invertible, so $$\det((A+tI)B-xI)=\det(B(A+tI)-xI).$$ Both sides of this equation are elements of the polynomial ring $\Bbb F[x,t]$. Now we can replace $t$ by $0$ to get $$\det(AB-xI)=\det(BA-xI).$$

Angina Seng
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