# Eigenspace/Kernel/Linearly independent/Section
## Lemma

Let $K$ be a field, $V$ a $K$-vector space and
    $\varphi \colon V\longrightarrow V$
a linear mapping. Then
    $\operatorname {ker} {\left(\varphi \right)}=\operatorname {Eig} _{0}{\left(\varphi \right)}\,.$
$0$
eigenvalue of $\varphi$ if and only if $\varphi$ is not

### Proof

$\Box$
More general, we have the following characterization.

## Lemma

Let $K$ be a field, $V$ a $K$-vector space and
    $\varphi \colon V\longrightarrow V$
a linear mapping. Let $\lambda \in K$. Then
    $\operatorname {Eig} _{\lambda }{\left(\varphi \right)}=\operatorname {ker} {\left(\lambda \cdot \operatorname {Id} _{V}-\varphi \right)}\,.$

### Proof

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

## Remark

Beside the eigenspace for $0\in K$, which is the kernel of the linear mapping, the eigenvalues $1$ and $-1$ are in particular interesting. The eigenspace for $1$ consists of all vectors which are sent to themselves. Restricted to this linear subspace, the mapping is just the identity, it is called the fixed space. The eigenspace for $-1$ 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 $K$ be a field, $V$ a $K$-vector space and
    $\varphi \colon V\longrightarrow V$
a linear mapping. Let $\lambda _{1}\neq \lambda _{2}$ be elements in $K$. Then
    $\operatorname {Eig} _{\lambda _{1}}{\left(\varphi \right)}\cap \operatorname {Eig} _{\lambda _{2}}{\left(\varphi \right)}=0\,.$

### Proof

$\Box$

## Lemma

Let $K$ be a field, $V$ a $K$-vector space and
    $\varphi \colon V\longrightarrow V$
a linear mapping. Let $v_{1},\ldots ,v_{n}$ be eigenvectors for (pairwise) different eigenvalues $\lambda _{1},\ldots ,\lambda _{n}\in K$. Then $v_{1},\ldots ,v_{n}$ are

### Proof

We prove the statement by induction on $n$. For $n=0$, the statement is true. Suppose now that the statement is true for less than $n$ vectors. We consider a representation of $0$, say
    $a_{1}v_{1}+\cdots +a_{n}v_{n}=0\,.$
We apply $\varphi$ to this and get, on one hand,
    $a_{1}\varphi (v_{1})+\cdots +a_{n}\varphi (v_{n})=\lambda _{1}a_{1}v_{1}+\cdots +\lambda _{n}a_{n}v_{n}=0\,.$
On the other hand, we multiply the equation with $\lambda _{n}$ and get
    $\lambda _{n}a_{1}v_{1}+\cdots +\lambda _{n}a_{n}v_{n}=0\,.$
We look at the difference of the two equations, and get
    $(\lambda _{n}-\lambda _{1})a_{1}v_{1}+\cdots +(\lambda _{n}-\lambda _{n-1})a_{n-1}v_{n-1}=0\,.$
By the induction hypothesis, we get for the coefficients $(\lambda _{n}-\lambda _{i})a_{i}=0$, $i=1,\ldots ,n-1$. Because of $\lambda _{n}-\lambda _{i}\neq 0$, we get $a_{i}=0$ for $i=1,\ldots ,n-1$, and because of $v_{n}\neq 0$, we also get $a_{n}=0$.
$\Box$

## Corollary

Let $K$ be a field, $V$ a finite-dimensional $K$-vector space and
    $\varphi \colon V\longrightarrow V$
a linear mapping. Then there exist at most $\dim _{\left(V\right)}$ many eigenvalues
$\varphi$

### Proof

$\Box$