# Endomorphism/Eigenspaces/Relations/Section

We have seen in fact that the eigenspace to $0$ is the kernel of the endomorphism. More general, the following characterization holds.

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

### 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 $\varphi (v)-\lambda v=0$ holds, which means ${\left(\varphi -\lambda \cdot \operatorname {Id} _{V}\right)}(v)=0$.
$\Box$
In particular, $\lambda \in K$ is an eigenvalue of $\varphi$ if and only if $\varphi -\lambda \operatorname {Id} _{V}$ is not injective. For a given $\lambda$, this property can be checked with the help of a linear system (or the determinant), and the eigenspace can be determined. However, it is not a linear problem to decide whether $\varphi$ has eigenvalues at all and how those can be determined. We will continue to study a linear mapping $\varphi \colon V\rightarrow V$ by considering the differences to homotheties $\varphi -\lambda \operatorname {Id} _{V}$ for various $\lambda$.
For an $n\times n$-matrix $M$, we have to determine the kernel of the matrix $M-\lambda E_{n}$. If, for example, we want t know whether the matrix ${\begin{pmatrix}4&7\\-2&5\end{pmatrix}}$ has the eigenvalue $3$, then
    ${\begin{pmatrix}4&7\\-2&5\end{pmatrix}}-3E_{2}={\begin{pmatrix}4&7\\-2&5\end{pmatrix}}-3{\begin{pmatrix}1&0\\0&1\end{pmatrix}}={\begin{pmatrix}1&7\\-2&2\end{pmatrix}}\,$
shows that this is not the case.

## 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

Let $v\in \operatorname {Eig} _{\lambda _{1}}{\left(\varphi \right)}\cap \operatorname {Eig} _{\lambda _{2}}{\left(\varphi \right)}$. Then
    $\lambda _{1}v=\varphi (v)=\lambda _{2}v\,.$
Therefore,
    $(\lambda _{1}-\lambda _{2})v=0\,,$
and this implies, because of $\lambda _{1}\neq \lambda _{2}$, $v=0$.
$\Box$

## 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 $\dim _{K}{\left(V\right)}$ many eigenvalues
$\varphi$

### Proof

$\Box$
In particular, an endomorphism on a finite-dimensional vector space has only finitely many eigenvalues.