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
asked 2021-02-15
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.

Answers (1)

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.

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