Let \(T\) be the linear operator on \(\RR^3\) defined by \[ T(x_1, x_2, x_3) = (x_1,\; x_3,\; -2x_2 - x_3). \] Let \(f\) be the polynomial over \(\RR\) defined by \(f = -x^3 + 2\). Find \(f(T)\).
Solution
Relative to the standard basis, \[ [T]= \begin{bmatrix} 1&0&0\\ 0&0&1\\ 0&-2&-1 \end{bmatrix}. \] A direct computation gives \[ [T]^3= \begin{bmatrix} 1&0&0\\ 0&2&-1\\ 0&2&3 \end{bmatrix}. \] Therefore \[ [f(T)]=- [T]^3+2I = \begin{bmatrix} 1&0&0\\ 0&0&1\\ 0&-2&-1 \end{bmatrix} =[T]. \] Thus \[ \boxed{f(T)=T}. \]