Endomorphism/Algebraic multiplicity/Introduction/Section

For a more detailed investigation of the eigenspaces, the following concept is useful.

Definition

Let

$\varphi \colon V\longrightarrow V$

be a linear mapping on a finite-dimensional ${}K$-vector space ${}V$, and ${}\lambda \in K$. The exponent of the linear polynomial ${}X-\lambda$ in the characteristic polynomial ${}\chi _{\varphi }$ is called the algebraic multiplicity of ${}\lambda$. It is denoted by

${}\mu _{\lambda }=\mu _{\lambda }(\varphi )\,.$

Recall that

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

is called the geometric multiplicity of ${}\lambda$. We know, due to fact, that one of these multiplicities is positive if and only if this is true for the other multiplicity, and this is the case if and only if ${}\lambda$ is an eigenvalue.

In general, the multiplicities can be different, we have, however, an estimate between them.

Lemma

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

$\varphi \colon V\longrightarrow V$

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

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

between the geometric and the

algebraic multiplicity.

Proof

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

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

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

$\Box$

Example

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

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

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

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

From

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

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