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

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

• This shows that $AB$ and $BA$ have different minimal polynomial. But characteristic polynomials are the same right? – 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 $(BA)^k=B(AB)^{k-1}A$.

• in general means for m not equal to n?? – Ri-Li Sep 16 '14 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. – 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 $$\begin{pmatrix}0 & 0 \\B & BA\end{pmatrix} \begin{pmatrix}I & -A \\0 & I\end{pmatrix}$$.