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

${}\operatorname {ker} {\left(\varphi \right)}=\operatorname {Eig} _{0}{\left(\varphi \right)}\,.$

In particular, ${}0$ is an

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

injective.

Proof

See exercise.

$\Box$

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

See exercise.

$\Box$

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

linearly independent.

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

for ${}\varphi$.

Proof

See exercise.

$\Box$