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

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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?

Answer 4

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*}

2. 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*}

3. 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.

Answer 5

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

Answer 6

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.

Answer 7

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

Answer 8

$\newcommand{\tr}{\operatorname{Tr}}$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.

Answer 9

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 $n <k$ 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 $\det (xI- ST)x^{n - m} = \det (x I - TS)$.

Answer 10

The following are two proofs by Ichiro Satake:

Answer 11

[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 $${ \det \begin{pmatrix} A &B \\ C &D \end{pmatrix} = \det(A) \det(\Delta _A) = \det(\Delta _D) \det(D) .}$$

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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.

Answer 12

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

Additional lensdump link: https://lensdump.com/i/sChcBM

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[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.

Answer 13

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. $\blacksquare$