Exercise 7.3.5

Let $f(t)=t^3-2t+t=t{(t-1)}^2$. Thus we have $f(T)=T_0$. So the minimal polynomial $p(t)$ must divide the polynomial $f(t)$. Since $T$ is diagonalizable, $p(t)$ could only be $t$, $(t-1)$, or $t(t-1)$. If $p(t)=t$, then $T=T_0$. If $p(t)=(t-1)$, then $T=I$. If $p(t)=t(t-1)$, then ${[T]}_{\beta }=\left(\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)$for some basis $\beta$.