Prove Theorem 10.8: In Theorem \(10.7\) (Problem 10.9), if \(f(t)\) is the minimal
polynomial of \(T\) (and \(g(t)\) and \(h(t)\) are monic), then \(g(t)\) is the
minimal polynomial of the restriction \(T_{1}\) of \(T\) to \(U\) and \(h(t)\) is the
minimal polynomial of the restriction \(T_{2}\) of \(T\) to \(W\).
Let \(m_{1}(t)\) and \(m_{2}(t)\) be the minimal polynomials of \(T_{1}\) and
\(T_{2}\), respectively. Note that \(g\left(T_{1}\right)=0\) and
\(h\left(T_{2}\right)=0\) because \(U=\) Ker \(g(T)\) and \(W=\) Ker \(h(T)\). Thus,
$$
m_{1}(t) \text { divides } g(t) \quad \text { and } \quad m_{2}(t) \text {
divides } h(t)
$$
By Problem 10.9, \(f(t)\) is the least common multiple of \(m_{1}(t)\) and
\(m_{2}(t) .\) But \(m_{1}(t)\) and \(m_{2}(t)\) are relatively prime because \(g(t)\)
and \(h(t)\) are relatively prime. Accordingly, \(f(t)=m_{1}(t) m_{2}(t)\). We
also have that \(f(t)=g(t) h(t)\). These two equations together with (1) and the
fact that all the polynomials are monic imply that \(g(t)=m_{1}(t)\) and
\(h(t)=m_{2}(t)\), as required.