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?

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    $\begingroup$ 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... $\endgroup$ – Qiaochu Yuan Feb 22 '13 at 21:12
  • $\begingroup$ @cmi obviously not. Try to figure out 2 different polynomials with the same set of roots. Is not hard. $\endgroup$ – Gaston Burrull Jul 4 '19 at 7:41
  • $\begingroup$ people.math.sc.edu/howard/Classes/700/charAB.pdf $\endgroup$ – Bach Aug 16 '19 at 19:32

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.

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    $\begingroup$ 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? $\endgroup$ – EuYu Feb 22 '13 at 17:37
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    $\begingroup$ @EuYu, I have no reference, sorry. $\endgroup$ – Nathan Portland Feb 22 '13 at 22:11
  • $\begingroup$ @EuYu one other method is the argument that matrices $A=(a_{ij})$ and $B=(b_{ij})$ are invertible over the field $K(a_{ij}, b_{ij})$ $\endgroup$ – Bananach Feb 15 '20 at 17:58
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    $\begingroup$ See this short note by JH Williamson from 1953 cambridge.org/core/services/aop-cambridge-core/content/view/… $\endgroup$ – Bananach Feb 15 '20 at 18:09

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.


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?

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    $\begingroup$ This shows that $AB$ and $BA$ have different minimal polynomial. But characteristic polynomials are the same right? $\endgroup$ – user464147 Nov 25 '17 at 1:39

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 $(BA)^k=B(AB)^{k-1}A$.

  • $\begingroup$ in general means for m not equal to n?? $\endgroup$ – Ri-Li Sep 16 '14 at 21:16
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    $\begingroup$ 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. $\endgroup$ – Marc van Leeuwen Dec 23 '15 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 $\chi_M(x) = det(x - M)$

Lets prove the fact: For square matrices $A$ and $B$ holds $det(AB - x) = det(BA - x) \Leftrightarrow \chi_{AB}(x) = \chi_{BA}(x)$.

If $det(A) \neq 0$ then it follows from $det(AB - x) = det(A^{-1}A)det(AB - x) = det(A^{-1})det(AB - x)det(A) = det(BA - x)$.

If $det(A) = 0$ there are finite number of such $s \in \mathbb R$ that $\chi_A(s)=0$ because $\chi_A(s)$ is a finite-degree polynomial. Then there are infinite number of such $s$ 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 $x$.

For square matrices we are done!

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


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.

$\det(xI-A) = \det((xI-A)^T) = \det(xI-A^T)$

2) Determinant of a block triangular matrix:

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

Consider $\begin{pmatrix}I & -A \\0 & I\end{pmatrix} \begin{pmatrix}AB & 0 \\B & 0\end{pmatrix}$ and $\begin{pmatrix}0 & 0 \\B & BA\end{pmatrix} \begin{pmatrix}I & -A \\0 & I\end{pmatrix}$.


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