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Mathematics LibreTexts

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:
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).
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 \(V\).