characteristic polynomial

A Problem in Linear Algebra

Bishal Deb

Sep 1, 2016 • 4 min read

In this article we shall show that the characteristic polynomial of both ${AB}$ and ${BA}$ are the same, where ${A}$ and ${B}$ are ${n\times n}$ matrices over ${R}$ , a ring with unity.

Let ${\Phi_T\in R[\lambda]}$ denote the characteristic polynomial of ${T}$ . Thus, we intend to show, ${\Phi_{AB}=\Phi_{BA}}$ .

We first take ${R=\mathbb{C}}$ and specialize in this ring. By multiplicity of the determinant function, we have that the characteristic polynomial of two similar matrices are the same.

We keep ${A}$ fixed and let ${B}$ be a diagonalizable (semisimple) matrix. Let ${T}$ be an invertible matrix such that ${TBT^{-1}}$ is diagonal. Then,

$\displaystyle \Phi_{AB}$

$\displaystyle = \Phi_{(TAT^{-1})(TBT^{-1})}$

$\displaystyle =det(\lambda I - (TAT^{-1})(TBT^{-1}))$

$\displaystyle =det((\lambda I - (TBT^{-1})(TAT^{-1}))^{T})$

$\displaystyle =\Phi_{(TBT^{-1})(TAT^{-1})}$

$\displaystyle =\Phi_{BA}$

Fact: The set of semisimple matrices is dense in ${M_n(\mathbb{C})}$ .

By making use of the this and continuity of det function we conclude that

$\displaystyle \Phi_{AB}=\Phi_{BA},$

$\displaystyle \forall A,B\in M_n(\mathbb{C}).$

Let ${X = [x_{ij}]_{n\times n}}$ and ${Y=[y_{ij}]_{n\times n}}$ be two matrices where ${\{x_{ij}\}}$ and ${\{y_{ij}\}}$ are ${2n^2}$ variables. Then,

$\displaystyle \Phi_{XY}-\Phi_{YX}\in\mathbb{C}[\{x_{ij}\},\{y_{ij}\},\lambda].$

We call this polynomial as ${P}$ . Corresponding to this polynomial we have a function ${\tilde{P} : \mathbb{C}^{2n^2+1}\rightarrow\mathbb{C}}$ where,

$\displaystyle \tilde{P}(\{a_{ij}\},\{b_{ij}\},l)=det(lI-[a_{ij}][b_ij])-det(lI-[a_{ij}][b_ij])$

As ${\Phi_{AB}=\Phi_{BA}}$ , we have that ${\tilde{P}}$ is the zero function. Thus, we have that ${P}$ is the zero polynomial.

$\displaystyle \mathbb{Z}[\{x_{ij}\},\{y_{ij}\},\lambda]\subset \mathbb{C}[\{x_{ij}\},\{y_{ij}\},\lambda]$

the identity, ${P=0}$ holds for entries in ${\mathbb{Z}}$ .

We make use of the following facts : –

There is a unique homomorphism from ${\mathbb{Z}\rightarrow R}$ where ${R}$ is a ring with unity. If ${\phi: R\rightarrow R'}$ is a homomorphism then there is a unique homomorphism ${\Phi:R[x_1,\ldots,x_n]\rightarrow R'}$ which preserves action of ${\phi}$ on constants and ${x_i\mapsto \alpha_i'\in R'}$ .