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(\lambda \cdot \operatorname {Id} _{V}-\varphi \right)}\,.

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

In particular, ${}\lambda \in K$ is an eigenvalue of ${}\varphi$ if and only if ${}\lambda \operatorname {Id} _{V}-\varphi$ 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 ${}\lambda \operatorname {Id} _{V}-\varphi$ for various ${}\lambda$ .

For an ${}n\times n$ -matrix ${}M$ , we have to determine the kernel of the matrix ${}\lambda E_{n}-M$ . If, for example, we want t know whether the matrix ${}{\begin{pmatrix}4&7\\-2&5\end{pmatrix}}$ has the eigenvalue ${}3$ , then

{}3E_{2}-{\begin{pmatrix}4&7\\-2&5\end{pmatrix}}=3{\begin{pmatrix}1&0\\0&1\end{pmatrix}}-{\begin{pmatrix}4&7\\-2&5\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

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 ${}a_{n}=0$ .