# Algebraic and Geometric Multiplicities

Theorem: For a square matrix A, the geometric multiplicity of its any eigenvalue is less than or equal to its algebraic multiplicity. ■

Let λ be an eigenvalue of a n×n matrix A, and suppose that the dimension of its eigenspace, ker(λI - A), is k. Let x1, x2, ... , xk be a basis for this eigenspace. We build n×k matrix X from these eigenvectors:
${\bf X} = \left[ {\bf x}_1 \ {\bf x}_2 \ \cdots \ {\bf x}_k \right] \qquad \Longrightarrow \qquad {\bf A}\,{\bf X} = \left[ {\bf A}\,{\bf x}_1 \ {\bf A}\,{\bf x}_2 \ \cdots \ {\bf A}\,{\bf x}_k \right] = \left[ \lambda{\bf x}_1 \ \lambda{\bf x}_2 \ \cdots \ \lambda{\bf x}_k \right] = \lambda \,{\bf X} .$
We complete matrix X with n×(n-k) matrix X' so that the square n×n matrix S = [X X'] becomes invertible. Then
$\left[ {\bf S}^{-1} {\bf X}\ {\bf S}^{-1} {\bf X}' \right] = {\bf S}^{-1} {\bf S} = \begin{bmatrix} {\bf I}_{k\times k} & {\bf 0} \\ {\bf 0} & {\bf I}_{(n-k)\times (n-k)} \end{bmatrix} ,$
so
${\bf S}^{-1} {\bf X} = \begin{bmatrix} {\bf I}_{k\times k} \\ {\bf 0} \end{bmatrix} .$
Compute
${\bf S}^{-1} {\bf A}\ {\bf S} = {\bf S}^{-1} \left[ {\bf S}^{-1} {\bf X}\ {\bf S}^{-1} {\bf X}' \right] = {\bf S}^{-1} \left[ \lambda {\bf X}\ {\bf A}\, {\bf X}' \right] = \left[ \lambda {\bf S}^{-1} {\bf X} \ {\bf S}^{-1}{\bf A}\, {\bf X}' \right] = \begin{bmatrix} \lambda {\bf I}_{k\times k} & {\bf \ast} \\ {\bf 0} & {\bf C}_{(n-k)\times (n-k)} \end{bmatrix} ,$
for some (n-k)×(n-k) matrix C. Since similar matrices have the same characteristic polynomial, we get
$\chi_{A} (z) = \chi_{S^{-1} A\,S} (z) = \chi_{\lambda\,I_k} (z) \, \chi_{C} (z) = \left( z- \lambda \right)^k \chi_{C} (z) .$
Consequenly, λ is a root of χA(z) = 0 with multiplicity at least k. ■