 ## 31 Aug A Problem in Linear Algebra

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.

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

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 ${x_i\mapsto \alpha_i'\in R'}$.

Using the above facts there is a unique homomorphism $\displaystyle \psi : \mathbb{Z}[\{x_{ij}\},\{y_{ij}\},\lambda]\rightarrow R[\lambda]$

which sends $\displaystyle x_{ij}\mapsto a_{ij}$ $\displaystyle y_{ij}\mapsto b_{ij}$ $\displaystyle \lambda \mapsto \lambda$

Thus, ${\psi(P)=det(\lambda I - [a_{ij}][b_{ij}]) -det(\lambda I - [b_{ij}][a_{ij}])}$.

But as ${P=0}$ in ${\mathbb{Z}[\{x_{ij}\},\{y_{ij}\},\lambda]}$ $\displaystyle \Phi_{AB}=\Phi_{BA}$

where ${A=[a_{ij}]}$.

Download this post as PDF (will not include images and mathematical symbols).

,
##### 1Comment
• ##### Ravi Hirapra
Posted at 23:00h, 01 September Reply

For a simple and easy proof visit exercise prob 8 and 9 of section 6.2 of Hoffman and Kunze ‘ linear algebra’