# Linear mapping/Multiplicities/Introduction/Section

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

${\displaystyle \varphi \colon V\longrightarrow V}$

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

${\displaystyle \dim _{}{\left(\operatorname {Eig} _{\lambda }{\left(\varphi \right)}\right)},}$

is called the geometric multiplicity of ${\displaystyle {}\lambda }$. 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 ${\displaystyle {}K}$ denote a field, and let ${\displaystyle {}V}$ denote a finite-dimensional vector space. Let

${\displaystyle \varphi \colon V\longrightarrow V}$

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

${\displaystyle {}\dim _{}{\left(\operatorname {Eig} _{\lambda }{\left(\varphi \right)}\right)}\leq \mu _{\lambda }(\varphi )\,}$

between the geometric and the algebraic multiplicity.

### Proof

Let ${\displaystyle {}m=\dim _{}{\left(\operatorname {Eig} _{\lambda }{\left(\varphi \right)}\right)}}$ and let ${\displaystyle {}v_{1},\ldots ,v_{m}}$ be a basis of this eigenspace. We complement this basis with ${\displaystyle {}w_{1},\ldots ,w_{n-m}}$ to get a basis of ${\displaystyle {}V}$, using fact. With respect to this basis, the describing matrix has the form

${\displaystyle {\begin{pmatrix}\lambda E_{m}&B\\0&C\end{pmatrix}}.}$

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

${\displaystyle \Box }$

## Example

We consider the ${\displaystyle {}2\times 2}$-shearing matrix

${\displaystyle {}M={\begin{pmatrix}1&a\\0&1\end{pmatrix}}\,,}$

with ${\displaystyle {}a\in K}$. The characteristic polynomial is

${\displaystyle {}\chi _{M}=(X-1)(X-1)\,,}$

so that ${\displaystyle {}1}$ is the only eigenvalue of ${\displaystyle {}M}$. The corresponding eigenspace is

${\displaystyle {}\operatorname {Eig} _{1}{\left(M\right)}=\operatorname {ker} {\left({\begin{pmatrix}0&-a\\0&0\end{pmatrix}}\right)}\,.}$

From

${\displaystyle {}{\begin{pmatrix}0&-a\\0&0\end{pmatrix}}{\begin{pmatrix}r\\s\end{pmatrix}}={\begin{pmatrix}-as\\0\end{pmatrix}}\,,}$

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