eigenvalues of AB and of BA are identical

hbghlyj

Weinstein–Aronszajn identity：對任意兩方陣 $ A,B $，其特徵多項式 $ p_{AB}(t)=p_{BA}(t) $.
How can we prove Sylvester's determinant identity?
ab, ba, and the spectrum
當 $A$ 是非奇異矩陣，$AB$ 和 $BA$ 相似：$ {\displaystyle BA=A^{-1}(AB)A.} $
當$A$和$B$都是奇異矩陣，$AB$和$BA$不一定相似。反例：取
$$A = \left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right], B = \left[ \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} \right].$$
有 $AB = A$ 但 $BA = 0$。

hbghlyj

AA^T and A^TA share non-zero eigenvalues$\def\tr{\operatorname{Tr}}$

In fact, nonzero eigenvalues $AB$ and $BA$ are the same for any rectangular matrices
$A$ and $B$. This follows from the fact that $\tr((AB)^k) = \tr((BA)^k)$ and the coefficients of the characteristic polynomials of a square matrix $A$ are a function of $\tr(A^k).$

$\tr((AB)^k) = \tr((BA)^k)$ follows from cyclic property of trace
the coefficients of $\chi_A$ are a function of $\tr(A^k)$ follows from the Faddeev–LeVerrier algorithm

hbghlyj

The eigenvalues of AB and of BA are identical for all square A and B
page 2

hbghlyj

Linear Algebra, Hoffman & Kuntze, page 190
How would you solve this tantalizing Halmos problem?

Czhang271828

恒等式 $\lambda^n\det (\lambda I_m-BA)=\lambda^m\det (\lambda I_n-AB)$ 出现过很多次了, 最近一次是此贴.