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 \( \alpha_1 , \alpha_2 , \ldots , \alpha_n . \) ■