A Problem in Linear Algebra

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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'}.

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}]}.

I am a masters student at Chennai Mathematical Institute. I am also an editor of the English section.

  • 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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