 # Let Z be a cycle of generalized eigenvectors of a linear operator T on V that corresponds to the eingenvalue lambda.Prove that span(Z) is a T-invariant subspace of V. Yulia 2021-02-15 Answered
Let Z be a cycle of generalized eigenvectors of a linear operator T on V that corresponds to the eingenvalue lambda.Prove that span(Z) is a T-invariant subspace of V.
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Here, we are given that Z is the cycle of generalized eigenvectors of a linear operator T on V that corresponds to the eigenvalue $\lambda$.
Let T be a linear operator on a vector space , and let be an eigenvalue of . Suppose that
$Z={v}_{1},{v}_{2},\dots ,{v}_{n}$
Or if p is the smallest positive integer such that ${\left(T-\lambda I\right)}^{p}v=0,Z=\left\{{\left(T-\lambda I\right)}^{p-1}v,{\left(T-\lambda I\right)}^{p-2}v,\dots ,\left(T-\lambda I\right)v,v\right\}$,
cycles of generalized eigenvectors of corresponding to .
Step 2
Now, we need to show that span(Z) is a T-invariant subspace of V.
Let us take a vector w from span(Z).
So, we can be written as linear combination of ${v}_{1},{v}_{2},\dots ,{v}_{n}$.
i.e., .
Now, it can be observed that if we can show that the vectors are T invariant then w is also T invariant.
$T\left({v}_{i}\right)=T\left({\left(T-\lambda I\right)}^{k}v\right)$, for some k (by the definition)
$=\left(T-\lambda I+\lambda I\right)\left({\left(T-\lambda I\right)}^{k}v\right)$
$=\left(T-\lambda I\right){\left(T-\lambda I\right)}^{k}v+\lambda I{\left(T-\lambda I\right)}^{k}v$
$={\left(T-\lambda I\right)}^{k+1}v+\lambda I{\left(T-\lambda I\right)}^{k}v$
$={\left(T-\lambda I\right)}^{k+1}v+{\left(T-\lambda I\right)}^{k}\left(\lambda v\right)$
And note that ${\left(T-\lambda I\right)}^{k+1}v\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\left(T-\lambda I\right)}^{k}\left(\lambda v\right)$ are also in span(Z), i.e., T-invariant.
Now, since w was taken randomly, span(Z) is a T-invariant subspace of V.

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