Characteristic polynomials of AB and BA

Posted: May 15, 2010 in Elementary Algebra; Problems & Solutions, Linear Algebra
Tags: ,

Let \mathbb{M}_{r,s}(\mathbb{C}) be the set of all r \times s matrices with entries from \mathbb{C}. Let I be the identity matrix and f_D(x) the characteristic polynomial of a square matrix D.

Problem. Let 1 \leq m \leq n \in \mathbb{N} and suppose that A \in \mathbb{M}_{m,n}(\mathbb{C}) and B \in \mathbb{M}_{n,m}(\mathbb{C}). Prove that f_{BA}(x)=x^{n-m}f_{AB}(x).

Solution. We’ll consider three cases:

Case 1. n=m and A is invertible : in this case we’ll have BA=A^{-1}ABA. Thus AB and BA are similar and hence f_{BA}(x)=f_{AB}(x).

Case 2. n=m and A is not invertible : in this case, since the equation \det(tI + A) = 0 has a finitely many solutions for t \in \mathbb{R}, there exists some \epsilon > 0 such that \det(tI+A) \neq 0, for all 0 < |t|< \epsilon, i.e. tI + A is invertible for all 0 < |t| < \epsilon. Thus, by Case 1, for any t with 0 < |t| < \epsilon, we have

\det(xI - (tI+A)B)=\det(xI - B(tI+A)).

Now letting t \to 0 we’ll get f_{AB}(x)=\det(xI - AB)=\det(xI - BA)=f_{BA}(x).

Case 3. n > m : in this case we add n-m zero rows (resp. columns) to A (resp. B) to get an n \times n matrix A_0 (resp. B_0). It follows immediately that

B_0A_0=BA and A_0B_0=\begin{pmatrix} AB & 0 \\ 0 & 0 \end{pmatrix}.


f_{A_0B_0}(x)=\det(xI - A_0B_0)=x^{n-m} \det(xI - AB)=x^{n-m}f_{AB}(x).

Thus,  by Case 1 and Case 2

f_{BA}(x)=f_{B_0A_0}(x)=f_{A_0B_0}(x)=x^{n-m}f_{AB}(x). \ \Box

Remark. So if n=m, then, counting multiplicity, the eigenvlaues of AB and BA are the same. If n>m, counting multiplicity, the eigenvalues of BA are the same as the eigenvalues of AB plus n-m eigenvalues which are all 0.

  1. pidi says:

    Case2, you already assume that 0<t0′ ?

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