Products of matrices in either order have the same characteristic polynomial

Question

Let $A, B$ be square matrices over $\Bbb C$. Prove that matrices $AB$ and $BA$ have the same characteristic polynomial.

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I know it's a famous problem and found various answers. However, I am at my first year of math degree and my knowledge is very limited.

I have never seen matrix which the entires of the matrix is matrices themselves. We never spoke in the class about limits of matrices (those the sort of solutions I saw online).

So, this question is kind of "challenge" for us to prove with our basic linear algebra knowledge. If any one knows a solution (complicated as it may be as long as it dosent require more than the basic knowledge) it would help a lot. thank you very much

Answer 1

I provide here a detailed (and to some extent elementary) explanation of this proof. Note that we assume the (fairly elementary) fact that for square matrices $A,B$, we have $\det(AB) = \det(A)\det(B)$.

With that, we proceed as follows. Let $n$ be the size of the matrices $A$ and $B$.

$\det(A)$ is a polynomial on the entries of $A$. For example, when $A$ is $2 \times 2$, we have $$ \det A = a_{11}a_{22} - a_{12}a_{21}. $$ The above expression for $\det(A)$ is a "polynomial" in that it requires only addition and multiplication.

Note: In general, a similar expression can be attained with the Leibniz expansion of the determinant. Just as the polynomial $f(x,y) = x^2 - 2y^2 + xy$ is a polynomial on two variables, so is the determinant of $A$ a polynoimal on $n^2$ variables.

Similarly, $p_1 = \det(xI - AB)$ and $p_2 = \det(x I - BA)$ are polynomials on the entries $a_{ij}$ of $A$, the entries $b_{ij}$ of $B$, and $x$. Our goal is to show that $p_1 = p_2$. The key to the proof is the following: $$ \begin{align} \det(A)\cdot p_1 &= \det(xI - AB)\det(A) = \det([xI - AB]A) = \det(xA - ABA) \\ & = \det(A[x I - BA]) = \det(A) \det(x I - BA) = \det(A)\cdot p_2. \end{align} $$ Now, it suffices to show that the following is true.

Claim: Suppose that $p_1,p_2,q$ are non-zero polynomials on $m$ variables such that $$ q(x_1,\dots,x_m)p_1(x_1,\dots,x_m) = q(x_1,\dots,x_m)p_2(x_1,\dots,x_m). $$ Then it must hold that $p_1 = p_2$.

Proof: Let $p = p_1 - p_2$. It is equivalent to show that if $q\cdot p = 0$, then $p = 0$. In other words, it suffices to show that $\Bbb C[x_1,\dots,x_m]$ is an integral domain, as is done on this post.

Answer 2

Here is a classical solution.

Step 1 If $B$ is invertible.

Then \begin{align}P_{AB}(x)&= \det(xI-AB)\\&=\det(xB^{-1}B-AB)\\&= \det(xB^{-1}-A) \det(B)\\&=\det(B) \det(xB^{-1}-A)\\& = \det(xI-BA)\end{align}

Step 2 The general case.Let $B$ be arbitrary.

We want to show that $$\det(\lambda I-AB)=\det(\lambda I -BA)$$ for all $\lambda$.

Fix an arbitrary $\lambda$. Define $$P(x)= \det(\lambda I-A(B-xI))-\det(\lambda I -(B-xI)A)$$ This is a polynomial in $x$ of degree at most $n$. Moreover, by Step 1, we have $P(x)=0$ for all $x$ which are not eigenvalues of $B$. Therefore, $P$ has infinitely many roots and hence $$P=0$$

In particular $P(0)=0$ which shows the claim.

Answer 3

Here's a simple proof that assumes OP knows how to telescope a finite geometric series and take a limit of a (complex) scalar sequence.

main idea is observing:
$\text{trace}\Big(\big(AB\big)^k\Big)=\text{trace}\Big(\big(BA\big)^k\Big)$ for $k \in\{1,2,3,...\}$

and the below proves that if traces of two $\text{n x n}$ complex matrices match for all powers of $k$, then the two matrices have the same eigenvalues. (With a lot more work the below can be developed into Newton's Identities though that seemed outside the scope.)

with $AB$ having distinct eigenvalues $\lambda_{AB}=\{\lambda_1, \lambda_2, ..., \lambda_d\}$ and $BA$ having distinct eigenvalues $\lambda_{BA}=\{\gamma_1, \gamma_2, ..., \gamma_r\}$. Between the two sets, there is a maximum modulus eigenvalue denoted $\sigma$. And for purposes of ordering, in each case the ordering is from smallest modulus to largest.

$x \in \mathbb C, \big \vert x\big \vert \gt \sigma$
i.e. $x$ may be any value on the complex plane outside the circle with radius $\sigma$ (that is centered at the origin)

$\sum_{k=1}^d\alpha_k\frac{1-(\frac{\lambda_k}{x})^m}{1-\frac{\lambda_k}{x}} =n + \sum_{k=1}^{m-1} \text{trace}\Big(\big(\frac{1}{x}AB\big)^k\Big) =n + \sum_{k=1}^{m-1} \text{trace}\Big(\big(\frac{1}{x}BA\big)^k\Big)=\sum_{j=1}^r\beta_j\frac{1-(\frac{\gamma_j}{x})^m}{1-\frac{\gamma_j}{x}}$

where $\alpha_k$ and $\beta_j$ are positive integers, i.e. denoting the algebraic multiplicities of the eigenvalues. Taking limits as $m\to \infty$

$\sum_{k=1}^d\alpha_k\frac{1}{1-\frac{\lambda_k}{x}}= \sum_{j=1}^r\beta_j\frac{1}{1-\frac{\gamma_j}{x}}$
suppose WLOG that $\vert \lambda_d\vert =\sigma$ and multiply each side by $(1-\frac{\lambda_d}{x})$

$LHS=(1-\frac{\lambda_d}{x})\sum_{k=1}^d\alpha_k\frac{1}{1-\frac{\lambda_k}{x}}= (1-\frac{\lambda_d}{x})\sum_{j=1}^r\beta_j\frac{1}{1-\frac{\gamma_j}{x}}=RHS$

take a limit as $x\to \lambda_d$ e.g. sequentially $x_t = (1 +\frac{1}{t})\lambda_d$

$0\lt \alpha_d= \lim_{x\to \lambda_d}\sum_{j=1}^r\beta_j(1-\frac{\lambda_d}{x})\frac{1}{1-\frac{\gamma_j}{x}}$
but $\lim_{x\to \lambda_d}(1-\frac{\lambda_d}{x})\frac{1}{1-\frac{\gamma_j}{x}}= 1$ iff $\gamma_j = \lambda_d$ and 0 otherwise.
Thus there is a maximum modulus eigenvalue in common with the same algebraic multiplicities and the labeling (other than for modulus purposes) is arbitrary so assume WLOG that $\gamma_r = \lambda_d$. Thus $\beta_r = \alpha_d$. Now recurse on the strictly smaller subproblem

$LHS=\sum_{k=1}^{d-1}\alpha_k\frac{1}{1-\frac{\lambda_k}{x}}= \sum_{j=1}^{r-1}\beta_j\frac{1}{1-\frac{\gamma_j}{x}}=RHS$