Endomorphism/Geometric multiplicity/Section

The restriction of a linear mapping to an eigenspace is the homothety with the corresponding eigenvalue, thus a very simple linear mapping. For a diagonal matrix

${\begin{pmatrix}d_{1}&0&\cdots &\cdots &0\\0&d_{2}&0&\cdots &0\\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&d_{n-1}&0\\0&\cdots &\cdots &0&d_{n}\end{pmatrix}},$

the standard basis has the property that every basis vector is an eigenvector for the linear mapping given by the matrix. In this case, it is easy to describe the eigenspaces, see example; the eigenspace for ${}d$ consists of all linear combinations of the standard vectors ${}e_{i}$, for which ${}d_{i}$ equals ${}d$. In particular, the dimension of the eigenspace equals the number how often ${}d$ occurs as an diagonal element. In general, the dimensions of the eigenspaces are important invariants of an endomorphism.

Definition

Let ${}K$ be a field, ${}V$ a ${}K$-vector space and

$\varphi \colon V\longrightarrow V$

a linear mapping. For ${}\lambda \in K$ we call

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

the geometric multiplicity

of ${}\lambda$.

In particular, a number ${}\lambda \in K$ is an eigenvalue of ${}\varphi$ if and only if its geometric multiplicity is at least ${}1$. It is easy to give examples where the geometric multiplicity of an eigenvalue is any number between ${}1$ and the dimension of the space.

Lemma

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

$\varphi \colon V\longrightarrow V$

be a linear mapping. Then the sum of the eigenspaces ${}\operatorname {Eig} _{\lambda }{\left(\varphi \right)}$ is direct, and we have

${}\sum _{\lambda \in K}\dim _{K}{\left(\operatorname {Eig} _{\lambda }{\left(\varphi \right)}\right)}\leq \dim _{K}{\left(V\right)}\,.$

Proof

This follows directly from fact.

$\Box$