Skip to main content
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

# 12.3: Eigenspaces

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 $$V$$.