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,


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,


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.


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]


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

  • 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’

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