12.3: Eigenspaces
- Page ID
- 2079
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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 \(V\).