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T is an operator on $R ^3$ with M(T) $\left( \begin {matrix} 1&1&0 \\ 0&1&0\\0&0&3 \end{matrix} \right)$ with respect to the standard basis.

This is the answer from the book : the subspace of generalised eigenvectors for the eigenvalue 3 is span{(0,0,1)}

I have no idea how they worked this out.

I always thought since 3 only appears once then it can have at most 1 generalized eigenvector but this is obviously not right.

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    $\begingroup$ This matrix is in Jordan normal form, and $(0,0,1)$ is an eigenvector for the eigenvalue $3$. In the present case, the generalised eigenspace is simply the eigenspace. $\endgroup$
    – Bernard
    Nov 14, 2015 at 12:27
  • $\begingroup$ Do you have a formal definition of "generalized eigenvector" to work with? All vectors of the form $(0,0,t)$ with $t\ne 0$ are actual eigenvectors for the eigenvalue $3$. $\endgroup$ Nov 14, 2015 at 12:27
  • $\begingroup$ So far all I know is $(T-EI)^j$v=0. I have no idea what a generalized eigenvector is, why it is useful, how we get them etc. $\endgroup$
    – user289419
    Nov 14, 2015 at 12:40

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