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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$\begingroup$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.
$\endgroup$ 1 $\begingroup$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.
$\endgroup$ 4 $\begingroup$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?
$\endgroup$ 1 $\begingroup$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 $(BA)^k=B(AB)^{k-1}A$.
$\endgroup$ 2 $\begingroup$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.
$\endgroup$ $\begingroup$Yes, $AB$ and $BA$ have the same characteristic polynomial.
Basic facts: $\det(A^T) = \det(A)$, $\det(AB) = \det(A) \det(B)$
- $A$ and $A^T$ share the same characteristic polynomial.
\begin{align*} \det(xI-A) = \det((xI-A)^T) = \det(xI-A^T) \end{align*}
- 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*}
- Determinant of a block triangular matrix:
\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.
$\endgroup$ $\begingroup$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 $\chi_{BA}(x) = \chi_{B'A'}(x) = \chi_{A'B'}(x) = x^{n-m}\chi_{AB}(x)$
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