Prove Lemma 10.13: Let \(T: V \rightarrow V\) be a linear operator whose minimal
polynomial is \(f(t)^{n}\) where \(f(t)\) is a monic irreducible polynomial. Then
\(V\) is the direct sum of \(T\) -cyclic subspaces \(Z_{i}=Z\left(v_{i}, T\right),
i=1, \ldots, r,\) with corresponding \(T\) -annihilators
\\[
f(t)^{n_{1}}, f(t)^{n_{2}}, \ldots, f(t)^{n_{r}}, \quad n=n_{1} \geq n_{2}
\geq \cdots \geq n_{r}
\\]
Any other decomposition of \(V\) into the direct sum of \(T\) -cyclic subspaces
has the same number of components and the same set of \(T\) -annihilators.
The proof is by induction on the dimension of \(V\). If \(\operatorname{dim}
V=1\), then \(V\) is \(T\) -cyclic and the lemma holds. Now suppose dim \(V>1\) and
that the lemma holds for those vector spaces of dimension less than that of
\(V\)
Because the minimal polynomial of \(T\) is \(f(t)^{n}\), there exists \(v_{1} \in
V\) such that \(f(T)^{n-1}\left(v_{1}\right) \neq 0 ;\) hence, the \(T\)
-annihilator of \(v_{1}\) is \(f(t)^{n}\). Let \(Z_{1}=Z\left(v_{1}, T\right)\) and
recall that \(Z_{1}\) is \(T\) -invariant. Let \(V=V / Z_{1}\) and let \(T\) be the
linear operator on \(\bar{V}\) induced by \(T\). By Theorem 10.16 , the minimal
polynomial of \(\bar{T}\) divides \(f(t)^{n}\); hence, the hypothesis holds for
\(\bar{V}\) and \(\bar{T}\). Consequently, by induction, \(\bar{V}\) is the direct
sum of \(\bar{T}\) -cyclic subspaces; say,
\\[
\bar{V}=Z\left(\bar{v}_{2}, \bar{T}\right) \oplus \cdots \oplus
Z\left(\bar{v}_{r}, \bar{T}\right)
\\]
where the corresponding \(\bar{T}\) -annihilators are \(f(t)^{\text {th }}
\sqrt[7]{2}, \ldots, f(t)^{n_{7}}, n \geq n_{2} \geq \cdots \geq n_{r_{i}}\) We
claim that there is a vector \(v_{2}\) in the coset \(\bar{v}_{2}\) whose \(T\)
-annihilator is \(\bar{f}(t)^{m_{2}},\) the \(\bar{T}\) -annihilator of
\(\bar{v}_{2}\). Let \(w\) be any vector in \(\bar{v}_{2}\). Then \(f(T)^{n_{2}}(w)
\in Z_{1}\). Hence, there exists a polynomial \(g(t)\) for which
\\[
f(T)^{n_{2}}(w)=g(T)\left(v_{1}\right)
\\]
Because \(f(t)^{n}\) is the minimal polynomial of \(T,\) we have, by (1)
\\[
0=f(T)^{n}(w)=f(T)^{n-n_{2}} g(T)\left(v_{1}\right)
\\]
But \(f(t)^{n}\) is the \(T\) -annihilator of \(v_{1} ;\) hence, \(f(t)^{n}\) divides
\(f(t)^{n-n_{2}} g(t),\) and \(\operatorname{so~} g(t)=f(t)^{n} h(t)\) for some
polynomial \(h(t)\). We set
\\[
v_{2}=w-h(T)\left(v_{1}\right)
\\]
Because \(w-v_{2}=h(T)\left(v_{1}\right) \in Z_{1}, v_{2}\) also belongs to the
coset \(\bar{v}_{2}\). Thus, the \(T\) -annihilator of \(v_{2}\) is a multiple of
the \(T\) -annihilator of \(\bar{v}_{2}\). On the other hand, by (1)
\\[
f(T)^{n_{2}}\left(v_{2}\right)=f(T)^{n_{*}}\left(w-h(T)\left(v_{1}\right)\right)=f(T)^{n_{2}}(w)-g(T)\left(v_{1}\right)=0
\\]
Consequently, the \(T\) -annihilator of \(v_{2}\) is \(f(t)^{n_{2}}\), as claimed.
Similarly, there exist vectors \(v_{3}, \ldots, v_{r} \in V\) such that \(v_{i}
\in \bar{v}_{i}\) and that the \(T\) -annihilator of \(v_{i}\) is \(f(t)^{n_{i}}\)
the \(\vec{T}\) -annihilator of \(\bar{v}_{i} .\) We set
\\[
Z_{2}=Z\left(v_{2}, T\right), \quad \ldots, \quad Z_{r}=Z\left(v_{r}, T\right)
\\]
Let \(d\) denote the degree of \(f(t),\) so that \(f(t)^{n_{i}}\) has degree \(d
n_{l} .\) Then, because \(f(t)^{n_{i}}\) is both the \(T\) -annihilator of \(v_{i}\)
and the \(\bar{T}\) -annihilator of \(\bar{v}_{i},\) we know that
\\[
\left\\{v_{i}, T\left(v_{i}\right), \ldots, T^{\ln
_{i}-1}\left(v_{i}\right)\right\\} \quad \text { and }
\quad\left\\{\vec{v}_{i} \cdot \vec{T}\left(\bar{v}_{i}\right), \ldots,
\bar{T}^{\operatorname{dr}_{i}-1}\left(\bar{v}_{i}\right)\right\\}
\\]
are bases for \(Z\left(v_{i}, T\right)\) and \(Z(\overline{v_{i}}, \bar{T}),\)
respectively, for \(i=2, \ldots, r .\) But \(\bar{V}=Z(\overline{v_{2}}, T)
\oplus \cdots \oplus Z(\overline{v_{r}}, T)\)
hence,
\\[
\left(\bar{v}_{2}\right), \ldots, \bar{v}_{r}, \ldots, T^{d
n_{r}-1}\left(\bar{v}_{r}\right.
\\]
is a basis for \(V\). Therefore, by Problem 10.26 and the relation
\(\vec{T}^{\prime}(\vec{v})=\overline{T^{i}(v)}\) (see Problem 10.27 ),
\\[
\left\\{v_{1}, \ldots, T^{d n_{1}-1}\left(v_{1}\right), v_{2}, \ldots,
T^{m_{2}-1}\left(v_{2}\right), \ldots, v_{r}, \ldots, T^{d
n_{r}-1}\left(v_{r}\right)\right\\}
\\]
is a basis for \(V .\) Thus, by Theorem \(10.4, V=Z\left(v_{1}, T\right) \oplus
\cdots \oplus Z\left(v_{r}, T\right),\) as required. It remains to show that
the exponents \(n_{1}, \ldots, n_{r}\) are uniquely determined by \(T .\) Because
\(d=\operatorname{degree}\) of \(f(t)\)
\\[
\operatorname{dim} V=d\left(n_{1}+\cdots+n_{r}\right) \quad \text { and }
\quad \operatorname{dim} Z_{i}=d n_{i}, \quad i=1, \ldots, r
\\]
Also, if \(s\) is any positive integer, then (Problem 10.59 )
\(f(T)^{3}\left(Z_{i}\right)\) is a cyclic subspace generated by
\(f(T)^{s}\left(v_{i}\right),\) and it has dimension \(d\left(n_{i}-s\right)\) if
\(n_{i}>s\) and dimension 0 if \(n_{i} \leq s\)
Now any vector \(v \in V\) can be written uniquely in the form
\(v=w_{1}+\cdots+w_{r},\) where \(w_{i} \in Z_{i}\) Hence, any vector in
\(f(T)^{s}(V)\) can be written uniquely in the form
\\[
f(T)^{s}(v)=f(T)^{s}\left(w_{1}\right)+\cdots+f(T)^{s}\left(w_{r}\right)
\\]
where \(f(T)^{s}\left(w_{i}\right) \in f(T)^{s}\left(Z_{i}\right) .\) Let \(t\) be
the integer, dependent on \(s,\) for which
\\[
n_{1}>s, \quad \ldots, \quad n_{t}>s, \quad n_{t+1} \geq s
\\]
Then
\\[
f(T)^{s}(V)=f(T)^{s}\left(Z_{1}\right) \oplus \cdots \oplus
f(T)^{s}\left(Z_{t}\right)
\\]
and so
\\[
\operatorname{dim}\left[f(T)^{s}(V)\right]=d\left[\left(n_{1}-s\right)+\cdots+\left(n_{t}-s\right)\right]
\\]
The numbers on the left of (2) are uniquely determined by \(T\). Set \(s=n-1\),
and (2) determines the number of \(n_{i}\) equal to \(n\). Next set \(s=n-2,\) and
(2) determines the number of \(n_{i}\) (if any) equal to \(n-1\). We repeat the
process until we set \(s=0\) and determine the number of \(n_{i}\) equal to \(1 .\)
Thus, the \(n_{l}\) are uniquely determined by \(T\) and \(V\), and the lemma is
proved.
Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine 
