Do matrices $AB$ and $BA$ have the same minimal and characteristic polynomials?

Let $A, B$ be two square matrices of order $n$. Do $AB$ and $BA$ have same minimal and characteristic polynomials?

I have a proof only if $A$ or $B$ is invertible. Is it true for all cases?

• The coefficients of the characteristic polynomial are continuous functions in the entries of a matrix, so if the characteristic polynomials of $AB$ and $BA$ coincide for a dense set of $A$ (or a dense set of $B$) then they always coincide. The coefficients of the minimal polynomial, on the other hand... Commented Feb 22, 2013 at 21:12
• @cmi obviously not. Try to figure out 2 different polynomials with the same set of roots. Is not hard. Commented Jul 4, 2019 at 7:41
• people.math.sc.edu/howard/Classes/700/charAB.pdf
– Bach
Commented Aug 16, 2019 at 19:32

Before proving $AB$ and $BA$ have the same characteristic polynomials show that if $A_{m\times n}$ and $B_{n\times m}$ then characteristic polynomials of $AB$ and $BA$ satisfy following statement: $$x^n|xI_m-AB|=x^m|xI_n-BA|$$ therefore easily conclude if $m=n$ then $AB$ and $BA$ have the same characteristic polynomials.

Define $$C = \begin{bmatrix} xI_m & A \\B & I_n \end{bmatrix},\ D = \begin{bmatrix} I_m & 0 \\-B & xI_n \end{bmatrix}.$$ We have \begin{align*} \det CD &= x^n|xI_m-AB|,\\ \det DC &= x^m|xI_n-BA|. \end{align*} and we know $\det CD=\det DC$ if $m=n$ then $AB$ and $BA$ have the same characteristic polynomials.

• I understood the proof, it's nice.. but is there any intuition abt why we consider C and D in such way? Commented May 8, 2021 at 7:00
• @Believer Maybe some intuition is that right multiplication gives column operation, and left multiplication give left multiplication. So $CD$ is related to giving the thing related to $AB$, and $DC$ gives $BA$. This also means that $E=\begin{pmatrix} I_m & -A\\ 0 & xI_n\end{pmatrix}$ would also work for the purpose of the proof. (Also you are trying to do row or column reduction to make $C$ triangular, hence the choice of $-B$ or $-A$ at the corresponding positions.) Commented Feb 23 at 16:22

If $A$ is invertible then $A^{-1}(AB)A= BA$, so $AB$ and $BA$ are similar, which implies (but is stronger than) $AB$ and $BA$ have the same minimal polynomial and the same characteristic polynomial. The same goes if $B$ is invertible.

In general, from the above observation, it is not too difficult to show that $AB$, and $BA$ have the same characteristic polynomial, the type of proof could depends on the field considered for the coefficient of your matrices though. If the matrices are in $\mathcal{M}_n(\mathbb C)$, you use the fact that $\operatorname{GL}_n(\mathbb C)$ is dense in $\mathcal{M}_n(\mathbb C)$ and the continuity of the function which maps a matrix to its characteristic polynomial. There are at least 5 other ways to proceed (especially for other field than $\mathbb C$).

In general $AB$ and $BA$ do not have the same minimal polynomial. I'll let you search a bit for a counter example.

• 5 other ways? I'm quite curious what those ways are. I only know of the continuity argument and an argument involving determinant identities on block matrices. Would it be possible to provide a reference to some other methods?
– EuYu
Commented Feb 22, 2013 at 17:37
• @EuYu, I have no reference, sorry. Commented Feb 22, 2013 at 22:11
• @EuYu one other method is the argument that matrices $A=(a_{ij})$ and $B=(b_{ij})$ are invertible over the field $K(a_{ij}, b_{ij})$ Commented Feb 15, 2020 at 17:58
• See this short note by JH Williamson from 1953 cambridge.org/core/services/aop-cambridge-core/content/view/… Commented Feb 15, 2020 at 18:09

Hint: Consider $A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$. What do you get in that case?

• This shows that $AB$ and $BA$ have different minimal polynomial. But characteristic polynomials are the same right?
– user464147
Commented Nov 25, 2017 at 1:39

Yes, $$AB$$ and $$BA$$ have the same characteristic polynomial.

Basic facts: $$\det(A^T) = \det(A)$$, $$\det(AB) = \det(A) \det(B)$$

1. $$A$$ and $$A^T$$ share the same characteristic polynomial.

\begin{align*} \det(xI-A) = \det((xI-A)^T) = \det(xI-A^T) \end{align*}

1. Similar matrices have the same characteristic polynomial. If $$B = PAP^{-1}$$,

\begin{align*} \det(xI - B) &= \det(xI - PAP^{-1}) \\ &= \det(P(xI - A)P^{-1}) \\ &= \det(P)\det(xI - A)\det(P^{-1}) \\ &= \det(xI - A) \end{align*}

1. Determinant of a block triangular matrix (A special case of Schur's formula):

\begin{align*} \det \begin{pmatrix}A & B \\0 & C\end{pmatrix} = \det(A) \det(C) \end{align*}

Using block multiplication, please verify that $$\begin{pmatrix}I & -A \\0 & I\end{pmatrix} \begin{pmatrix}AB & 0 \\B & 0\end{pmatrix} = \begin{pmatrix}0 & 0 \\B & BA\end{pmatrix} \begin{pmatrix}I & -A \\0 & I\end{pmatrix}$$.

Therefore, the matrices $$\begin{pmatrix}AB & 0 \\B & 0\end{pmatrix}$$ and $$\begin{pmatrix}0 & 0 \\B & BA\end{pmatrix}$$ are similar, and have the same characteristic polynomial.

\begin{align*} \det\left[x\begin{pmatrix}I & 0 \\0 & I\end{pmatrix} - \begin{pmatrix}AB & 0 \\B & 0\end{pmatrix}\right] &= \det(xI - AB) \det(xI) \end{align*} \begin{align*} \det\left[x\begin{pmatrix}I & 0 \\0 & I\end{pmatrix} - \begin{pmatrix}0 & 0 \\B & BA\end{pmatrix}\right] &= \det(xI) \det(xI - BA) \end{align*}

And there it is. But $$AB$$ and $$BA$$ do not need to have the same minimal polynomial. See Jim's answer for a counterexample.

It's not true that their characteristic polynomials will be the same in the general case. The best result in this general vein is the following.

Let $A\in\mathbb{F}^{m \times n}$ and let $B\in\mathbb{F}^{n \times m}$, and $AB$, $BA$ with minimal polynomials (over $\mathbb{F}$) $m_{AB}(x)$ and $m_{BA}(x)$ respectively. Then one of the following holds:

$m_{AB}(x) = m_{BA}(x)$, or $m_{AB}(x) = x \cdot m_{BA}(x)$, or $x\cdot m_{AB}(x) = m_{BA}(x)$.

It's easy, just use the fact that $(BA)^k=B(AB)^{k-1}A$.

• in general means for m not equal to n?? Commented Sep 16, 2014 at 21:16
• The question says (and always has said) square matrices $A,B$. If you are answering a more general question, then you should announce this. Also, answering a more general question is only useful if this is no more difficult than the actual question, or if the more general solution sheds more light on the solution. Commented Dec 23, 2015 at 21:10

For square matrices, the characteristic polynomials are same, but for $$A$$ a matrix of size $$m \times n$$ and $$B$$ a matrix of size $$n \times m$$ we have $$x^{m}C_{BA}(x)=x^{n}C_{AB}(x)$$. This implies that the nonzero eigenvalue of $$AB$$, counted with multiplicities, are same as nonzero eigenvalue of $$BA$$.

That is, if $$A$$ is of size 7×4 and $$B$$ is of size 4×7 and assume that the 4×4 matrix $$BA$$ has nonzero eigenvalues 1,1,3 so fourth eigenvalue of $$BA$$ is 0. Then the 7×7 matrix $$AB$$ will also have nonzero eigenvalue 1,1,3 and remaining four eigenvalue of $$AB$$ are zero.

There are a lot of proofs for characteristic polynomials to be same. I want to provide mine. It may be more complicated, but it is less "consider magic product of matrices".

Let $$\chi_M(x)$$ denotes a characteristic polynomial $$\chi_M(x) = \det(x - M)$$.

For square matrices $$A$$ and $$B$$ we have $$\det(AB - x) = \det(BA - x) \Leftrightarrow \chi_{AB}(x) = \chi_{BA}(x)$$. We prove this by considering separately the two cases $$\det(A)=0$$ and $$\det(A)\neq0$$:

1. If $$\det(A) \neq 0$$, then the statement follows from $$\det(AB - x) = \det(A^{-1}A)\det(AB - x) \\= \det(A^{-1})\det(AB - x)\det(A) = \det(BA - x).$$
2. If $$\det(A) = 0$$, there are finite number of $$s \in \mathbb R$$ such that $$\chi_A(s)=0$$, because $$\chi_A(s)$$ is a finite-degree polynomial. Then there are infinitely many $$s$$ such 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 $$x$$.

This proves the statement for squared matrices.

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 $$\chi_{BA}(x) = \chi_{B'A'}(x) = \chi_{A'B'}(x) = x^{n-m}\chi_{AB}(x)$$

• @user264745 Александр Тряпицын's proof of 2. is correct: For every fixed $x$, he said he views $\chi_{(A-s)B}(x)$ and $\chi_{B(A-s)}(x)$ as polynomials in $s$ which coincide on an infinite set, and concludes they are equal. The only trouble with this proof is that is works only on an infinite field. But this can be repaired by considering the infinite field "universal" for this situation: the field $K$ of rational fractions with rational coefficients and $2n^2$ indeterminates $X_{i,j},Y_{i,j}$, and the universal matrices $A,B\in M_n(K)$ whose entries are these indeterminates. Commented Jan 12, 2023 at 11:21
• I was about to type that "universal proof" as a new answer, but I discovered it was already given here by @anon Commented Jan 12, 2023 at 11:44
• @AnneBauval Unfortunately “universal proof” is far from my reach. I don’t know anything about field extension. Commented Jan 12, 2023 at 12:03
• @user264745 I wanted to clean up this section and I thought my previous comment was sufficient. But if you take me by flattery, ok I shall rewrite it. And most other comments can be deleted please Commented Jan 12, 2023 at 15:39
• If $P(X),Q(X)\in\Bbb R[X]$ are such that $P(s)=Q(s)$ for infinitely many values of $s$ (or only for more than $\deg(P-Q)$ values) then $P(X)=Q(X)$ (i.e. their coefficients are equal) hence $P(s)=Q(s)$ for every $s.$ Александр Тряпицын applies this to $P(X)=\chi_{(A-X)B}(x),Q(X)=\chi_{B(A-X)}(x)$ for $x$ fixed. Commented Jan 12, 2023 at 15:45

$$\newcommand{\tr}{\operatorname{Tr}}$$I’d like to add a more computational perspective to the elegant, more analytic proofs given here.

Because we can reconstruct the characteristic polynomial from the traces, it actually suffices to show that $$\tr(AB)=\tr(BA),\tr((AB)^2)=\tr((BA)^2),\cdots,\tr((AB)^n)=\tr((BA)^n)$$ to say that $$\chi_{AB}=\chi_{BA}$$ for two $$n$$-square matrices $$A,B$$.

But those trace equalities are easy and well known. $$\tr(AB)=\tr(BA)$$ follows from straightforward computation, and then (say) to show: $$\tr((AB)^4)=\tr(ABABABAB)=\tr(BABABABA)=\tr((BA)^4)$$All you have to do is use a transposition: if $$X:=A$$ and $$Y:=BABABAB$$ then we want to show $$\tr(XY)=\tr(YX)$$, which is true!

This should work for matrices over any integral domain.

• No, this argument does not work for matrices over any ring. You need $1,2,\ldots,n$ to be non-zero-divisors in order to be able to reconstruct the characteristic polynomial from the traces. Commented Aug 16, 2023 at 12:16
• You did in the last sentence of this answer, even though the question indeed didn't :) Commented Aug 16, 2023 at 12:21
• oh my apologies I forgot I ever wrote that Commented Aug 16, 2023 at 12:25
• @darijgrinberg my bad. I have made a hopefully acceptable fix Commented Aug 16, 2023 at 12:25
• With small modification this argument gives the result over any commutative ring. The argument should be: work over polynomial ring $\mathbb Z[\mathbf x]$-- Newton's Identities work fine here. We conclude that the sum of $r\times r$ principal minors of $(AB)$ minus the sum of $r\times r$ principal minors of $(BA)$ is the zero polynomial. This survives when we do a substitution homomorphism $\phi: \mathbb Z[\mathbf x]\longrightarrow R$ for arbitrary commutative ring $R$, and allows us to conclude $AB$ and $BA$ with components in $R$ have the same characteristic polynomials. Commented Jan 8 at 17:44

Here is another proof, based on exterior algebras. (This should go through for any free finitely generated module $$M, N$$ over a commutative ring $$A$$, but not too sure.)

Lemma. Let $$T$$ be an endomorphism over $$M$$, with $$\dim_A M = m$$. $$\det (xI - T) = \sum_{k = 0}^\infty \text{Tr} (\bigwedge^k(T)) (-1)^k x^{m-k}$$

This can be found in e.g. N. Bourbaki, Algebra I, Chapter 3, $$\S$$8, no. 5, Proposition 11. Now $$\bigwedge^k$$ commutes with composition (i.e. a functor), so we may say for $$A$$-linear maps $$T: M \to N$$ and $$S : N \to M$$,

$$\text{Tr}\bigwedge^k(TS) = \text{Tr}\bigwedge^k(T) \bigwedge^k(S) = \text{Tr}\bigwedge^k(S)\bigwedge^k(T) = \text{Tr} \bigwedge^k(ST)$$

Now we can prove the theorem: we have $$\det (xI - ST) = \sum_{k = 0}^\infty \text{Tr}(\bigwedge^k(TS))(-1)^k x^{m - k}$$

Consider the case $$\dim_A N = n \geq m$$: we multiply by $$x^{n - m}$$, whence we have $$\det (xI - ST)x^{n-m} = \det(xI - TS)$$ When $$n < m$$, all terms with $$n vanishes since $$\bigwedge^k (N) = \{0\}$$. Thus we have

$$\det (xI - ST) = x^{m - n} \sum_{k = 0}^n\text{Tr}(\bigwedge^k(TS))(-1)^k x^{n - k} = x^{m - n} \det (xI - TS)$$

In all cases, we have proved $$\det (xI- ST)x^{n - m} = \det (x I - TS)$$.

• "whence we have" --> "whence we need to show". Commented Aug 16, 2023 at 13:19
• Nice proof, although it doesn't quite generalize to finitely generated modules. Finitely generated projective modules of constant rank should do, probably. Maybe we can do away with the constant rank condition if we pass to the "reverse characteristic polynomial" $\det\left(I-xA\right)$. Commented Aug 16, 2023 at 13:21

The following are two proofs by Ichiro Satake:

[The identity also follows naturally on taking Schur complements]

Schur complements:

Consider a block matrix $${ S = \begin{pmatrix} A &B \\ C &D \end{pmatrix} ,}$$ where $${ A }$$ is $${ p \times p }$$ and $${ D }$$ is $${ q \times q .}$$
Assuming $${ A }$$ is invertible, the deletion of block $${ C }$$ by elementary row operations (this doesn’t disturb block $${ A }$$) looks like { \begin{align*} \begin{pmatrix} A &B \\ C &D \end{pmatrix} &\to \begin{pmatrix} I &O \\ -CA ^{-1} &I \end{pmatrix} \begin{pmatrix} A &B \\ C &D \end{pmatrix} \\ &= \begin{pmatrix} A &B \\ O &D - CA ^{-1} B \end{pmatrix}. \end{align*} } Further deletion of block $${ B }$$ by elementary column operations (this doesn’t disturb block $${ A }$$) looks like { \begin{align*} \begin{pmatrix} A &B \\ O &D - C A ^{-1} B \end{pmatrix} &\to \begin{pmatrix} A &B \\ O &D - C A ^{-1} B \end{pmatrix} \begin{pmatrix} I &-A^{-1} B \\ O &I \end{pmatrix} \\ &= \begin{pmatrix} A &O \\ O &D - C A ^{-1} B \end{pmatrix}. \end{align*} } In combined form, $${ \begin{pmatrix} I &O \\ -CA ^{-1} &I \end{pmatrix} \begin{pmatrix} A &B \\ C &D \end{pmatrix} \begin{pmatrix} I &-A^{-1}B \\ O &I \end{pmatrix} = \begin{pmatrix} A &O \\ O &D-CA^{-1}B \end{pmatrix} . }$$ The bottom right block $${ \Delta _A := D - C A ^{-1} B }$$ which remains is called Schur complement of $${ A }$$ in $${ S .}$$

Similarly, assuming $${ D }$$ is invertible, the deletion of block $${ C }$$ by elementary column operations (this doesn’t disturb block $${ D }$$) and deletion of block $${ B }$$ by elementary row operations (this doesn’t disturb block $${ D }$$) gives $${ \begin{pmatrix} I &-BD^{-1} \\ O &I \end{pmatrix} \begin{pmatrix} A &B \\ C &D \end{pmatrix} \begin{pmatrix} I &O \\ -D^{-1}C &I \end{pmatrix} = \begin{pmatrix} A - BD^{-1}C &O \\ O &D \end{pmatrix}. }$$ The top left block $${ \Delta _D := A - B D^{-1} C }$$ which remains is called Schur complement of $${ D }$$ in $${ S .}$$

Since inverses $${ \begin{pmatrix} I &O \\ X &I \end{pmatrix} ^{-1} = \begin{pmatrix} I &O \\ -X &I \end{pmatrix}, \quad \begin{pmatrix} I &X \\ O &I \end{pmatrix} ^{-1} = \begin{pmatrix} I &-X \\ O &I \end{pmatrix}, }$$ the above equations can be rewritten as factorisations { \begin{align*} \begin{pmatrix} A &B \\ C &D \end{pmatrix} &= \begin{pmatrix} I &O \\ CA^{-1} &I \end{pmatrix} \begin{pmatrix} A &O \\ O &\Delta _A \end{pmatrix} \begin{pmatrix} I &A^{-1}B \\ O &I \end{pmatrix} \\ &= \begin{pmatrix} I &BD^{-1} \\ O &I \end{pmatrix} \begin{pmatrix} \Delta _D &O \\ O &D \end{pmatrix} \begin{pmatrix} I &O \\ D^{-1}C &I \end{pmatrix} . \end{align*} } Taking determinants gives $${ \det \begin{pmatrix} A &B \\ C &D \end{pmatrix} = \det(A) \det(\Delta _A) = \det(\Delta _D) \det(D) .}$$

Original question:

Let matrices $${ M \in \mathbb{F} ^{p \times q}, }$$ $${ N \in \mathbb{F} ^{q \times p} .}$$ Consider the block matrix $${ \begin{pmatrix} A &B \\ C &D \end{pmatrix} = \begin{pmatrix} tI &M \\ N &I \end{pmatrix}. }$$ It’s Schur complement factorisations are { \begin{align*} \begin{pmatrix} tI &M \\ N &I \end{pmatrix} &= \begin{pmatrix} I &O \\ t^{-1} N &I \end{pmatrix} \begin{pmatrix} tI &O \\ O &I - t^{-1} NM \end{pmatrix} \begin{pmatrix} I &t^{-1}M \\ O &I \end{pmatrix} \\ &= \begin{pmatrix} I &M \\ O &I \end{pmatrix} \begin{pmatrix} tI - MN &O \\ O &I \end{pmatrix} \begin{pmatrix} I &O \\ N &I \end{pmatrix} . \end{align*}} Taking determinants gives $${ t ^p \det(t^{-1}(tI - NM)) = \det(tI - MN) }$$ that is $${ t ^{p-q} \det(tI - NM) = \det(tI - MN) ,}$$ as needed.

[The result follows from $${ m = n }$$ case of the argument here]

[Same argument expanded]

Let $${ A \in \mathbb{F} ^{m \times n}, B \in \mathbb{F} ^{n \times m} }$$ with $${ m \leq n }.$$ Goal is to relate characteristic polynomials of $${ AB, BA }.$$

Consider the composite matrix $${ \begin{pmatrix} I _m &A \\ B &I _n \end{pmatrix}. }$$

Deleting $${ B }$$ using row operations gives $${ \begin{pmatrix} I _m &0 \\ -B &I _n\end{pmatrix} \begin{pmatrix} I _m &A \\ B &I _n \end{pmatrix} = \begin{pmatrix} I _m &A \\ 0 &{\color{green}{I _n - BA}}\end{pmatrix} }.$$Deleting $${ B }$$ using column operations gives $${ \begin{pmatrix} I _m &A \\ B &I _n \end{pmatrix} \begin{pmatrix} I _m &0 \\ -B &I _n \end{pmatrix} = \begin{pmatrix} {\color{purple}{I _m - AB}} &A \\ 0 &I _n \end{pmatrix}. }$$ To compare characteristic polynomials, one would want the green entry to be $${ {\color{green}{x I _n - BA}} }$$ and purple entry to be $${ {\color{purple}{x I _m - AB}} }$$ instead.
So the following modification of above two equations works: $${ \begin{pmatrix} I _m &0 \\ -B &{\color{green}{x}}I _n \end{pmatrix} \begin{pmatrix} {\color{purple}{x}}I _m &A \\ B &I _n \end{pmatrix} = \begin{pmatrix} x I _m &A \\ 0 &{\color{green}{x I _n - BA}}\end{pmatrix}, }$$ and $${ \begin{pmatrix} {\color{purple}{x}}I _m &A \\ B &I _n \end{pmatrix} \begin{pmatrix} I _m &0 \\ -B &{\color{green}{x}}I _n \end{pmatrix} = \begin{pmatrix} {\color{purple}{x I _m - AB}} &xA \\ 0 &xI _n \end{pmatrix} .}$$Taking determinants gives $${ x ^m \det({\color{green}{x I _n - BA}}) = x ^n \det({\color{purple}{x I _m - AB}}) }$$ as needed.

• Thanks for mentioning, here is an additional lensdump link. Commented Nov 27, 2023 at 11:27
• This is a duplicate of the currently highest upvoted answer in the thread. What new insight does this provide?
– Pedro
Commented Nov 27, 2023 at 12:39
• @Pedro It imho motivates the choice of block matrices (“Deleting B by row and column operations respectively”) and then proceeds to characteristic polynomial case Commented Nov 27, 2023 at 12:48
• Could you then please type this in LaTeX and expand on the new insight? Thank you! Image only posts are not great, in particular due to link rot.
– Pedro
Commented Nov 27, 2023 at 12:54

I intend to dumb down a bit modified version of the highly abstract proof presented by @Anon here.

Let $$\mathbb F$$ be an arbitrary field and we pick matrices $$A$$ and $$B$$ from $$\mathbb F^{n\times n}$$.

Case 1: $$B$$ is invertible.
Check that $$AB=B^{-1}(BA)B$$ and hence $$AB\sim BA$$. Now use the fact that similar matrices have the same characteristic polynomial.

Case 2: $$B$$ is not invertible.
What do we do? We will make $$B$$ invertible in one way or the other :)
Consider the polynomial ring $$\mathbb F[x_1, x_2, \ldots, x_{n^2}]$$. It's an integral domain so we can embed it in its quotient field $$\mathbb K=\mathbb F(x_1, x_2, \ldots, x_{n^2})$$ i.e., the field of rational functions in $$n^2$$ indeterminates (variables) over $$\mathbb F$$.
Now, we will construct an invertible matrix $$B' \in \mathbb K^{n\times n}$$.
Note that each entry of this matrix is a rational function. We define $$(i,j)$$-th entry of $$B'$$ to be $$x_{(i-1)n+j}$$. $$B'=\begin{pmatrix}x_1 & x_2 &\ldots & x_n\\ x_{n+1} & x_{n+2} & \ldots & x_{2n}\\ \vdots&\vdots&&\vdots\\ x_{(n-1)n+1} &\ldots&\ldots&x_{n^2}\end{pmatrix}\tag*{}$$ Clearly, rows of $$B'$$ are linearly independent because each entry is an indeterminate so there can't be a non-trivial linear combination of the rows which gives zero. In particular, $$\det(B')\neq 0$$ and $$B'$$ is invertible.
We can treat $$A$$ as a matrix in $$\mathbb K^{n\times n}$$ where each entry is a constant polynomial.
Using the result proven in case 1, we see that $$AB'$$ and $$B'A$$ have the same characteristic polynomial. It's just that we have a different field. The result still holds.
Thus, we have $$\det(xI-AB')=\det(xI-B'A)\tag{01}$$ where $$x$$ is a scalar in $$\mathbb K$$ and $$I$$ is the identity matrix in $$\mathbb K^{n\times n}$$.
Since polynomials are function and we have a polynomial in two different forms, we can evaluate either of them at certain point and obtain the same value.
If we evaluate the polynomials $$(01)$$ at $$x_1=B_{11}$$, $$x_2=B_{12}$$,$$\ldots$$, $$x_{n}=B_{1n}$$,$$\ldots$$, $$x_{n^2}=B_{nn}$$, we can replace $$B'$$ by $$B$$. This gives us: $$\det(xI-AB)=\det(xI-BA)\tag{02}$$

Hence, proved. $$\blacksquare$$