Linear mapping/Multiplicities/Introduction/Section

For a more detailed investigation of eigenspaces, the following concepts are necessary. Let

$\phi :V⟶V$

denote a linear mapping on a finite-dimensional vector space $V$, and $\lambda \in K$. Then the exponent of the linear polynomial $X-\lambda$ inside the characteristic polynomial ${\chi }_{\phi }$ is called the algebraic multiplicity of $\lambda$, symbolized as ${\mu }_{\lambda }:={\mu }_{\lambda }(\phi )$. The dimension of the corresponding eigenspace, that is

${\dim }_K\left({\mathrm{Eig}}_{\lambda }\left(\phi \right)\right),$

is called the geometric multiplicity of $\lambda$. Because of fact, the algebraic multiplicity is positive if and only if the geometric multiplicity is positive. In general, these multiplicities might be different, we have however always one estimate.

Lemma

Let $K$ denote a field, and let $V$ denote a finite-dimensional vector space. Let

$\phi :V⟶V$

denote a linear mapping and $\lambda \in K$. Then we have the estimate

${\dim }_K\left({\mathrm{Eig}}_{\lambda }\left(\phi \right)\right)\le {\mu }_{\lambda }(\phi )$

between the geometric and the

algebraic multiplicity.

Proof

Let $m={\dim }_K\left({\mathrm{Eig}}_{\lambda }\left(\phi \right)\right)$ and let $v_1,\dots ,v_m$ be a basis of this eigenspace. We complement this basis with $w_1,\dots ,w_{n-m}$ to get a basis of $V$, using fact. With respect to this basis, the describing matrix has the form

$\left(\begin{array}{cc}\lambda E_m& B\\ 0& C\end{array}\right).$

Ttherefore, the characteristic polynomial equals (using exercise) $(X-\lambda {)}^m\cdot {\chi }_C$, so that the algebraic multiplicity is at least $m$.

$◻$

Example

We consider the $2\times 2$-shearing matrix

$M=\left(\begin{array}{cc}1& a\\ 0& 1\end{array}\right),$

with $a\in K$. The characteristic polynomial is

${\chi }_M=(X-1)(X-1),$

so that $1$ is the only eigenvalue of $M$. The corresponding eigenspace is

${\mathrm{Eig}}_1\left(M\right)=\ker \left(\left(\begin{array}{cc}0& -a\\ 0& 0\end{array}\right)\right).$

From

$\left(\begin{array}{cc}0& -a\\ 0& 0\end{array}\right)\left(\begin{array}{c}r\\ s\end{array}\right)=\left(\begin{array}{c}-as\\ 0\end{array}\right),$

we get that $\left(\begin{array}{c}1\\ 0\end{array}\right)$ is an eigenvector, and in case $a\ne 0$, the eigenspace is one-dimensional (in case $a=0$, we have the identity and the eigenspace is two-dimensional). So in case $a\ne 0$, the algebraic multiplicity of the eigenvalue $1$ equals $2$, and the geometric multiplicity equals $1$.