Question # 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.

Upper Level Math
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. 2021-02-16

Here, we are given that Z is the cycle of generalized eigenvectors of a linear operator T on V that corresponds to the eigenvalue $$\displaystyle\lambda$$.
Let T be a linear operator on a vector space , and let be an eigenvalue of . Suppose that
$$\displaystyle{Z}={v}_{{1}},{v}_{{2}},\ldots,{v}_{{n}}$$
Or if p is the smallest positive integer such that $$\displaystyle{\left({T}-\lambda{I}\right)}^{{p}}{v}={0},{Z}={\left\lbrace{\left({T}-\lambda{I}\right)}^{{{p}-{1}}}{v},{\left({T}-\lambda{I}\right)}^{{{p}-{2}}}{v},\ldots,{\left({T}-\lambda{I}\right)}{v},{v}\right\rbrace}$$,
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 $$\displaystyle{v}_{{1}},{v}_{{2}},\ldots,{v}_{{n}}$$.
i.e., $$\displaystyle{w}=\alpha_{{1}}{v}_{{1}}+α_{{2}}{v}_{{2}}+\ldots+\alpha_{{n}}{v}_{{n}}\ {w}{h}{e}{r}{e}\ {a}{t}\ least\ {o}ne\ \alpha_{{i}}\ne{0}$$.
Now, it can be observed that if we can show that the vectors are T invariant then w is also T invariant.
$$\displaystyle{T}{\left({v}_{{i}}\right)}={T}{\left({\left({T}-\lambda{I}\right)}^{{k}}{v}\right)}$$, for some k (by the definition)
$$\displaystyle={\left({T}-\lambda{I}+\lambda{I}\right)}{\left({\left({T}-\lambda{I}\right)}^{{k}}{v}\right)}$$
$$\displaystyle={\left({T}-\lambda{I}\right)}{\left({T}-\lambda{I}\right)}^{{k}}{v}+\lambda{I}{\left({T}-\lambda{I}\right)}^{{k}}{v}$$
$$\displaystyle={\left({T}-\lambda{I}\right)}^{{{k}+{1}}}{v}+\lambda{I}{\left({T}-\lambda{I}\right)}^{{k}}{v}$$
$$\displaystyle={\left({T}-\lambda{I}\right)}^{{{k}+{1}}}{v}+{\left({T}-\lambda{I}\right)}^{{k}}{\left(\lambda{v}\right)}$$
And note that $$\displaystyle{\left({T}-\lambda{I}\right)}^{{{k}+{1}}}{v}{\quad\text{and}\quad}{\left({T}-\lambda{I}\right)}^{{k}}{\left(\lambda{v}\right)}$$ are also in span(Z), i.e., T-invariant.
Now, since w was taken randomly, span(Z) is a T-invariant subspace of V.