# Eigenspace/Kernel/Linearly independent/Section

## Lemma

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

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

a linear mapping. Then

${\displaystyle {}\operatorname {ker} {\left(\varphi \right)}=\operatorname {Eig} _{0}{\left(\varphi \right)}\,.}$
In particular, ${\displaystyle {}0}$ is an

eigenvalue of ${\displaystyle {}\varphi }$ if and only if ${\displaystyle {}\varphi }$ is not injective.

### Proof

${\displaystyle \Box }$

More general, we have the following characterization.

## Lemma

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

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

a linear mapping. Let ${\displaystyle {}\lambda \in K}$. Then

${\displaystyle {}\operatorname {Eig} _{\lambda }{\left(\varphi \right)}=\operatorname {ker} {\left(\lambda \cdot \operatorname {Id} _{V}-\varphi \right)}\,.}$

### Proof

Let ${\displaystyle {}v\in V}$. Then ${\displaystyle {}v\in \operatorname {Eig} _{\lambda }{\left(\varphi \right)}}$ if and only if ${\displaystyle {}\varphi (v)=\lambda v}$, and this is the case if and only if ${\displaystyle {}\lambda v-\varphi (v)=0}$ holds, which means ${\displaystyle {}{\left(\lambda \cdot \operatorname {Id} _{V}-\varphi \right)}(v)=0}$.

${\displaystyle \Box }$

## Remark

Beside the eigenspace for ${\displaystyle {}0\in K}$, which is the kernel of the linear mapping, the eigenvalues ${\displaystyle {}1}$ and ${\displaystyle {}-1}$ are in particular interesting. The eigenspace for ${\displaystyle {}1}$ consists of all vectors which are sent to themselves. Restricted to this linear subspace, the mapping is just the identity, it is called the fixspace. The eigenspace for ${\displaystyle {}-1}$ consists in all vector which are sent to their negative. On this linear subspace, the mapping acts like the reflection at the origin.

## Lemma

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

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

a linear mapping. Let ${\displaystyle {}\lambda _{1}\neq \lambda _{2}}$ be elements in ${\displaystyle {}K}$. Then

${\displaystyle {}\operatorname {Eig} _{\lambda _{1}}{\left(\varphi \right)}\cap \operatorname {Eig} _{\lambda _{2}}{\left(\varphi \right)}=0\,.}$

### Proof

${\displaystyle \Box }$

## Lemma

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

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

a linear mapping. Let ${\displaystyle {}v_{1},\ldots ,v_{n}}$ be eigenvectors for (pairwise) different eigenvalues ${\displaystyle {}\lambda _{1},\ldots ,\lambda _{n}\in K}$. Then ${\displaystyle {}v_{1},\ldots ,v_{n}}$ are linearly independent.

### Proof

We prove the statement by induction on ${\displaystyle {}n}$. For ${\displaystyle {}n=0}$, the statement is true. Suppose now that the statement is true for less than ${\displaystyle {}n}$ vectors. We consider a representation of ${\displaystyle {}0}$, say

${\displaystyle {}a_{1}v_{1}+\cdots +a_{n}v_{n}=0\,.}$

We apply ${\displaystyle {}\varphi }$ to this and get, on one hand,

${\displaystyle {}a_{1}\varphi (v_{1})+\cdots +a_{n}\varphi (v_{n})=\lambda _{1}a_{1}v_{1}+\cdots +\lambda _{n}a_{n}v_{n}=0\,.}$

On the other hand, we multiply the equation with ${\displaystyle {}\lambda _{n}}$ and get

${\displaystyle {}\lambda _{n}a_{1}v_{1}+\cdots +\lambda _{n}a_{n}v_{n}=0\,.}$

We look at the difference of the two equations, and get

${\displaystyle {}(\lambda _{n}-\lambda _{1})a_{1}v_{1}+\cdots +(\lambda _{n}-\lambda _{n-1})a_{n-1}v_{n-1}=0\,.}$

By the induction hypothesis, we get for the coefficients ${\displaystyle {}(\lambda _{n}-\lambda _{i})a_{i}=0}$, ${\displaystyle {}i=1,\ldots ,n-1}$. Because of ${\displaystyle {}\lambda _{n}-\lambda _{i}\neq 0}$, we get ${\displaystyle {}a_{i}=0}$ for ${\displaystyle {}i=1,\ldots ,n-1}$, and because of ${\displaystyle {}v_{n}\neq 0}$, we also get ${\displaystyle {}a_{n}=0}$.

${\displaystyle \Box }$

## Corollary

Let ${\displaystyle {}K}$ be a field, ${\displaystyle {}V}$ a finite-dimensional ${\displaystyle {}K}$-vector space and

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

a linear mapping. Then there exist at most ${\displaystyle {}\dim _{}{\left(V\right)}}$ many eigenvalues for ${\displaystyle {}\varphi }$.

### Proof

${\displaystyle \Box }$