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12.3: Eigenspaces

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fIn the previous example, we found two eigenvectors

$$\begin{pmatrix}-1\\1\\0\end{pmatrix} \mbox{ and }\begin{pmatrix}1\\0\\1\end{pmatrix}$$

for $$L$$, both with eigenvalue $$1$$. Notice that

$$\begin{pmatrix}-1\\1\\0\end{pmatrix} + \begin{pmatrix}1\\0\\1\end{pmatrix}=\begin{pmatrix}0\\1\\1\end{pmatrix}$$

is also an eigenvector of $$L$$ with eigenvalue $$1$$. In fact, any linear combination

$$r\begin{pmatrix}-1\\1\\0\end{pmatrix} + s\begin{pmatrix}1\\0\\1\end{pmatrix}$$

of these two eigenvectors will be another eigenvector with the same eigenvalue.

More generally, let $$\{ v_{1}, v_{2}, \ldots \}$$ be eigenvectors of some linear transformation $$L$$ with the same eigenvalue $$\lambda$$. A $$\textit{linear combination}$$ of the $$v_{i}$$ can be written $$c^{1}v_{1}+c^{2}v_{2}+\cdots$$ for some constants $$\{c^{1}, c^{2},\ldots \}$$. Then:

\begin{eqnarray*}
L(c^{1}v_{1}+c^{2}v_{2}+\cdots) &=& c^{1}Lv_{1}+c^{2}Lv_{2}+\cdots \textit{ by linearity of L}\\
&=& c^{1}\lambda v_{1}+c^{2}\lambda v_{2}+\cdots \textit{ since $$Lv_{i}=\lambda v_{i}$$ }\\
&=& \lambda (c^{1}v_{1}+c^{2}v_{2}+\cdots).
\end{eqnarray*}

So every linear combination of the $$v_{i}$$ is an eigenvector of $$L$$ with the same eigenvalue $$\lambda$$. In simple terms, any sum of eigenvectors is again an eigenvector $$\textit{if they share the same eigenvalue}$$.

The space of all vectors with eigenvalue $$\lambda$$ is called an $$\textit{eigenspace}$$. It is, in fact, a vector space contained within the larger vector space $$V$$: It contains $$0_{V}$$, since $$L0_{V}=0_{V}=\lambda 0_{V}$$, and is closed under addition and scalar multiplication by the above calculation. All other vector space properties are inherited from the fact that $$V$$ itself is a vector space. In other words, the subspace theorem, 9.1.1 chapter 9, ensures that $$V_{\lambda}:=\{v\in V|Lv=0\}$$ is a subspace of $$V$$.