So , I know that all eigenspaces of the linear operator are invariant subspaces for that operator. My question is : If we have ONE dimensional invariant subspace $L$ for linear operator $A$, is that $L$ subspace of eigenspace for $A$ or is that $L$ the whole eigenspace ? I know how to prove that $L$ is the subspace of eigenspace for $A$ , but is it the whole eigenspace ?
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$\begingroup$ It an be either... $\endgroup$– DonAntonioCommented Apr 22, 2017 at 19:41
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$\begingroup$ Just to be clear, you're asking: "If I have a one dimensional invariant subspace $L,$ is $L$ an eigenspace? And if $L$ is invariant, then does that mean $L$ is the entire eigenspace?" $\endgroup$– ChickenmancerCommented Apr 22, 2017 at 19:44
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$\begingroup$ onedimensional L , yes . $\endgroup$– domdragCommented Apr 22, 2017 at 19:45
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$\begingroup$ Certainly not: For the identity operator, the eigenspace is the whole vector space and all invariant subspaces are invariant, so for vector spaces of dimension $> 1$... $\endgroup$– Travis WillseCommented Apr 22, 2017 at 19:57
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Take for example
$$A=I=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}\;,\;\;L=\text{Span}\,\left\{\;\begin{pmatrix}(1,1,1)\end{pmatrix}\;\right\}$$
Thus $\;\dim L=1\;$ and it is $\;A\,-$ invariant, yet the whole eigenspace ( of the unique eigenvector $\;1\;$) is the whole $\;\Bbb R^3\;$ ...
answered Apr 22, 2017 at 19:56