Prove Theorem 10.12: Let \(Z(v, T)\) be a \(T\) -cyclic subspace, \(T_{v}\) the
restriction of \(T\) to \(Z(v, T),\) and \(m_{v}(t)=t^{k}+a_{k-1}
t^{k-1}+\cdots+a_{0}\) the \(T\) -annihilator of \(v .\) Then
(i) The set \(\left\\{v, T(v), \ldots, T^{k-1}(v)\right\\}\) is a basis of \(Z(v,
T) ;\) hence, \(\operatorname{dim} Z(v, T)=k\)
(ii) The minimal polynomial of \(T_{v}\) is \(m_{v}(t)\)
(iii) The matrix of \(T_{v}\) in the above basis is the companion matrix
\(C=C\left(m_{v}\right)\) of \(m_{v}(t)\) [which has 1 's below the diagonal, the
negative of the coefficients \(a_{0}, a_{1}, \ldots, a_{k-1}\) of \(m_{v}(t)\) in
the last column, and \(0 \text { 's elsewhere }]\).
(i) \(\quad\) By definition of \(m_{v}(t), T^{k}(v)\) is the first vector in the
sequence \(v, T(v), T^{2}(v), \ldots\) that, is a linear combination of those
vectors that precede it in the sequence; hence, the set \(B=\left\\{v, T(v),
\ldots, T^{k-1}(v)\right\\}\) is linearly independent. We now only have to show
that \(Z(v, T)=L(B)\), the linear span of \(B\). By the above, \(T^{k}(v) \in L(B)
.\) We prove by induction that \(T^{n}(v) \in L(B)\) for every \(n .\) Suppose
\(n>k\) and \(T^{n-1}(v) \in L(B)-\) that \(\quad\) is, \(\quad T^{n-1}(v) \quad\) is
\(\quad\) a \(\quad\) linear \(\quad\) combination \(\quad\) of \(\quad v, \ldots,
T^{k-1}(v) . \quad\) Then \(T^{n}(v)=T\left(T^{n-1}(v)\right)\) is a linear
combination of \(T(v), \ldots, T^{k}(v) .\) But \(T^{k}(v) \in L(B) ;\) hence,
\(T^{n}(v) \in L(B)\) for every \(n .\) Consequently, \(f(T)(v) \in L(B)\) for any
polynomial \(f(t) .\) Thus, \(Z(v, T)=L(B),\) and \(\operatorname{so~} B\) is a
basis, as claimed.
(ii) Suppose \(m(t)=t^{s}+b_{s-1} t^{s-1}+\cdots+b_{0}\) is the minimal
polynomial of \(T_{v} .\) Then, because \(v \in Z(v, T)\)
\\[
0=m\left(T_{v}\right)(v)=m(T)(v)=T^{s}(v)+b_{s-1} T^{s-1}(v)+\cdots+b_{0} v
\\]
Thus, \(T^{s}(v)\) is a linear combination of \(v, T(v), \ldots, T^{s-1}(v),\) and
therefore \(k \leq s\). However, \(m_{v}(T)=\mathbf{0}\) and so
\(m_{v}\left(T_{v}\right)=\mathbf{0} .\) Then \(m(t)\) divides \(m_{v}(t),\) and so
\(s \leq k .\) Accordingly, \(k=s\) and hence \(m_{v}(t)=m(t)\)
(iii)
\\[
\begin{array}{ll}
T_{v}(v) & =\quad T(v) \\
T_{v}(T(v)) & = \\
& = \\
T_{v}\left(T^{k-2}(v)\right) & = \\
T_{v}\left(T^{k-1}(v)\right) & =T^{k}(v)=-a_{0} v-a_{1} T(v)-a_{2}
T^{2}(v)-\cdots-a_{k-1} T^{k-1}(v)
\end{array}
\\]
By definition, the matrix of \(T_{v}\) in this basis is the tranpose of the
matrix of coefficients of the above system of equations; hence, it is \(C,\) as
required.
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