Answer to Question #143174 in Linear Algebra for Sourav Mondal

Question

Let T : R3 -> R3 be the linear operator defined by T(x1, x2, x3) = (x1, x3, -2x2- x3).

Let f(x) = - x³+ 2.

Find the operator f(T)

Answer 1

Given $T\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} x_1 \\ x_3 \\ -2x_2-x_3 \end{bmatrix}$

In matirx form we have $T\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} =\begin{bmatrix} 1 &0 &0 \\ 0 & 0& 1 \\ 0 & -2 & -1 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = AX$

Where $A=\begin{bmatrix} 1 &0 &0 \\ 0 & 0& 1 \\ 0 & -2 & -1 \end{bmatrix}$ ; $X = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$

and $f(x) = -x^3+2$

then $f(T)= f(A)=-A^3+2I$

$A^3=\begin{bmatrix} 1 &0 &0 \\ 0 & 2& -1 \\ 0 & 2 & 3 \end{bmatrix}$ , $2I =2\begin{bmatrix} 1 &0 &0 \\ 0 & 1& 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 &0 &0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$

$f(A)=\begin{bmatrix} -1 &0 &0 \\ 0 & -2 & 1 \\ 0 & -2 & -3 \end{bmatrix} + \begin{bmatrix} 2 &0 &0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} = \begin{bmatrix} 1 &0 &0 \\ 0 & 0& 1 \\ 0 & -2 & -1 \end{bmatrix} =A$

Therefore $f(T)=T$