## 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,

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

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 $\Phi_{AB}=\Phi_{BA}$, $\forall A,B\in M_n(\mathbb{C})$.

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,

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

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,

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

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

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

$\psi : \mathbb{Z}[\{x_{ij}\},\{y_{ij}\},\lambda]\rightarrow R[\lambda]$

which sends

$x_{ij}\mapsto a_{ij}$

$y_{ij}\mapsto b_{ij}$

$\lambda \mapsto \lambda$

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

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

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

where $A=[a_{ij}]$ and $B=[b_{ij}]$.

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#### Bishal Deb

Bishal Deb is an undergraduate student at Chennai Mathematical Institute. He is an editor of Gonit Sora.

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• ##### 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'