Eigenvalues of AB and BA

Consider a square matrix consisting of four blocks:

${\bf A} = \begin{bmatrix} {\bf I}_{k\times k} & {\bf 0}_{k \times (n-k)} \\ {\bf X}_{(n-k)\times k} & {\bf I}_{(n-k)\times (n-k)} \end{bmatrix} ,$
where X is (n-k)-by-k rectangular matrix, I are square identity matrices, and 0 is the ractangular zero $$k \times (n-k)$$ matrix. It can be verified by a computation that
$\begin{bmatrix} {\bf I}_{k\times k} & {\bf 0}_{k \times (n-k)} \\ -{\bf X}_{(n-k)\times k} & {\bf I}_{(n-k)\times (n-k)} \end{bmatrix} \begin{bmatrix} {\bf I}_{k\times k} & {\bf 0}_{k \times (n-k)} \\ {\bf X}_{(n-k)\times k} & {\bf I}_{(n-k)\times (n-k)} \end{bmatrix} = \begin{bmatrix} {\bf I}_{k\times k} & {\bf 0}_{k \times (n-k)} \\ {\bf 0}_{(n-k)\times k} & {\bf I}_{(n-k)\times (n-k)} \end{bmatrix} = {\bf I}_n ,$
which shows that the inverse of lower (upper) block triangular matrix is again a lower (upper) block triangular matrix.

Theorem: For arbitrary square $$m\times n$$ matrix A and $$n\times m$$ matrix B, and $$n\ge m$$ , the following statements are valid.

• The nonzero eigenvalues of $$m\times m$$ matrix AB and $$n\times n$$ matrix BA are the same, with the same algebraic multiplicities.
• If 0 is an eigenvalue of AB with algebraic multiplicity $$k\ge 0 ,$$ then 0 is an eigenvalues of BA with algebraic multiplicity k+n-m.
• If m=n, then the eigenvalues of AB and BA are the same, with the same algebraic multiplicities. ■

Let
${\bf X}_{(m+n)\times (m+n)} = \begin{bmatrix} \left( {\bf A} \,{\bf B} \right)_{m\times m} & {\bf A}_{m \times n} \\ {\bf 0}_{n\times m} & {\bf 0}_{n\times n} \end{bmatrix} \qquad\mbox{and} \qquad {\bf Y}_{(m+n)\times (m+n)} = \begin{bmatrix} {\bf 0}_{m\times m} & {\bf A}_{m \times n} \\ {\bf 0}_{n\times m} & \left( {\bf B} \,{\bf A} \right)_{n\times n} \end{bmatrix} .$
Then we perform the following similarity transformations:
\begin{align*} \begin{bmatrix} {\bf I}_{m\times n} & {\bf 0}_{m\times n} \\ {\bf B}_{n\times m} & {\bf 0}_{n\times n} \end{bmatrix} {\bf X}_{(m+n)\times (m+n)} \begin{bmatrix} {\bf I}_{m\times n} & {\bf 0}_{m\times n} \\ {\bf B}_{n\times m} & {\bf 0}_{n\times n} \end{bmatrix}^{-1} &= \begin{bmatrix} {\bf I}_{m\times n} & {\bf 0}_{m\times n} \\ {\bf B}_{n\times m} & {\bf 0}_{n\times n} \end{bmatrix} \begin{bmatrix} \left( {\bf A} \,{\bf B} \right)_{m\times m} & {\bf A}_{m \times n} \\ {\bf 0}_{n\times m} & {\bf 0}_{n\times n} \end{bmatrix} \begin{bmatrix} {\bf I}_{m\times n} & {\bf 0}_{m\times n} \\ {\bf B}_{n\times m} & {\bf 0}_{n\times n} \end{bmatrix}^{-1} \\ &= \begin{bmatrix} {\bf I}_{m\times n} & {\bf 0}_{m\times n} \\ {\bf B}_{n\times m} & {\bf 0}_{n\times n} \end{bmatrix} \begin{bmatrix} \left( {\bf A} \,{\bf B} \right)_{m\times m} & {\bf A}_{m \times n} \\ {\bf 0}_{n\times m} & {\bf 0}_{n\times n} \end{bmatrix} \begin{bmatrix} {\bf I}_{m\times n} & {\bf 0}_{m\times n} \\ -{\bf B}_{n\times m} & {\bf 0}_{n\times n} \end{bmatrix} \\ &= \begin{bmatrix} {\bf A}\,{\bf B} & {\bf A} \\ {\bf B}\,{\bf A}\,{\bf B} & {\bf B} \, {\bf A} \end{bmatrix} \begin{bmatrix} {\bf I}_{m\times n} & {\bf 0}_{m\times n} \\ -{\bf B}_{n\times m} & {\bf 0}_{n\times n} \end{bmatrix} \\ &= \begin{bmatrix} {\bf A}\,{\bf B} - {\bf A}\,{\bf B} & {\bf A} \\ {\bf B}\,{\bf A}\,{\bf B} - {\bf B}\,{\bf A}\,{\bf B} & {\bf B} \, {\bf A} \end{bmatrix} \\ &= \begin{bmatrix} {\bf 0}_{m\times m} & {\bf A}_{m \times n} \\ {\bf 0}_{n\times m} & \left( {\bf B} \,{\bf A} \right)_{n\times n} \end{bmatrix} = {\bf Y} . \end{align*}
Because X and Y are similar matrices, their characteristic polynomials are also equal, $$\chi_X (\lambda ) = \det \left( \lambda {\bf I}_{(m+n)\times (m+n)} - {\bf X} \right) = \chi_Y (\lambda ) = \det \left( \lambda {\bf I}_{(m+n)\times (m+n)} - {\bf Y} \right) .$$ Since
$\chi_X (\lambda ) = \chi_{AB} (\lambda ) \,\chi_{{\bf 0}_n} (\lambda ) = \lambda^n \chi_{AB} (\lambda )$
and
$\chi_Y (\lambda ) = \chi_{{\bf 0}_m} (\lambda ) \,\chi_{BA} (\lambda ) = \lambda^m \chi_{BA} (\lambda ) ,$
it follows that
$\chi_{BA} (\lambda ) = \det \left( \lambda {\bf I}_{(m+n)\times (m+n)} - {\bf B}\,{\bf A} \right) = \lambda^{n-m} \chi_{AB} (\lambda ) = \det \left( \lambda {\bf I}_{(m+n)\times (m+n)} - {\bf A}\,{\bf B} \right) .$
Hence, if $$\lambda_1 , \lambda_2 , \ldots , \lambda_m$$ are roots of $$\chi_{AB} (\lambda )$$ (the eigenvalues of AB), then n roots of χBA(λ)=0 (the eigenvalues of BA) are
$\lambda_1 , \lambda_2 , \ldots , \lambda_m , \underbrace{0, 0, \ldots , 0}_{n-m} .$
So we see that nonzero eigenvalues of both matrices AB and BA are identical, with the same multiplicies. However, if k is the multiplicity of zero as an eigenvalue of AB, its multiplicity as an eigenvalue of BA is k+n-m.
Example: The span of the empty set $$\varnothing$$ consists of a unique element 0. Therefore, $$\varnothing$$ is linearly independent and it is a basis for the trivial vector space consisting of the unique element---zero. Its dimension is zero.

Example: In $$\mathbb{R}^n ,$$ the vectors $$e_1 [1,0,0,\ldots , 0] , \quad e_2 =[0,1,0,\ldots , 0], \quad \ldots , e_n =[0,0,\ldots , 0,1]$$ form a basis for n-dimensional real space, and it is called the standard basis. Its dimension is n.

Example: Let us consider the set of all real $$m \times n$$ matrices, and let $${\bf M}_{i,j}$$ denote the matrix whose only nonzero entry is a 1 in the i-th row and j-th column. Then the set $${\bf M}_{i,j} \ : \ 1 \le i \le m , \ 1 \le j \le n$$ is a basis for the set of all such real matrices. Its dimension is mn.

Example: The set of monomials $$\left\{ 1, x, x^2 , \ldots , x^n \right\}$$ form a basis in the set of all polynomials of degree up to n. It has dimension n+1. ■

Example: The infinite set of monomials $$\left\{ 1, x, x^2 , \ldots , x^n , \ldots \right\}$$ form a basis in the set of all polynomials. ■

Theorem: Let V be a vector space and $$\beta = \left\{ {\bf u}_1 , {\bf u}_2 , \ldots , {\bf u}_n \right\}$$ be a subset of V. Then β is a basis for V if and only if each vector v in V can be uniquely decomposed into a linear combination of vectors in β, that is, can be uniquely expressed in the form

${\bf v} = \alpha_1 {\bf u}_1 + \alpha_2 {\bf u}_2 + \cdots + \alpha_n {\bf u}_n$
for unique scalars $$\alpha_1 , \alpha_2 , \ldots , \alpha_n .$$