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

\(\displaystyle{Z}={\left\lbrace{v}_{{1}},{v}_{{2}},\ldots,{v}_{{n}}\right\rbrace}\)

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., \(\displaystyle{w}=\alpha_{{1}}{v}_{{1}}+α{l}{p}{h}{a}_{{2}}{v}_{{2}}+\ldots+\alpha_{{n}}{v}_{{n}}{w}{h}{e}{r}{e}{a}{t}\le\ast{o}\ne\alpha_{{i}}\ne{0}\).

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.

Let T be a linear operator on a vector space , and let be an eigenvalue of . Suppose that

\(\displaystyle{Z}={\left\lbrace{v}_{{1}},{v}_{{2}},\ldots,{v}_{{n}}\right\rbrace}\)

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., \(\displaystyle{w}=\alpha_{{1}}{v}_{{1}}+α{l}{p}{h}{a}_{{2}}{v}_{{2}}+\ldots+\alpha_{{n}}{v}_{{n}}{w}{h}{e}{r}{e}{a}{t}\le\ast{o}\ne\alpha_{{i}}\ne{0}\).

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.