Answer to Question #106623 in Linear Algebra for Sourav mondal

Question

Let T : R3-> R3be 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

We have $f(T)=-T^3+2I$

$T\begin{pmatrix} x_1\\ x_2\\ x_3 \end{pmatrix}=\begin{pmatrix} 1&0&0\\ 0&0&1\\ 0&-2&-1 \end{pmatrix}\begin{pmatrix} x_1\\ x_2\\ x_3 \end{pmatrix}$

Then $f(T)\begin{pmatrix} x_1\\ x_2\\ x_3 \end{pmatrix}=(-T^3+2I)\begin{pmatrix} x_1\\ x_2\\ x_3 \end{pmatrix}=$

$=\left(-\begin{pmatrix} 1&0&0\\ 0&0&1\\ 0&-2&-1 \end{pmatrix}^3+\begin{pmatrix} 2&0&0\\ 0&2&0\\ 0&0&2 \end{pmatrix}\right)\begin{pmatrix} x_1\\ x_2\\ x_3 \end{pmatrix}$

$\begin{pmatrix} 1&0&0\\ 0&0&1\\ 0&-2&-1 \end{pmatrix}^2=\begin{pmatrix} 1&0&0\\ 0&0&1\\ 0&-2&-1 \end{pmatrix}\begin{pmatrix} 1&0&0\\ 0&0&1\\ 0&-2&-1 \end{pmatrix}=$

$=\begin{pmatrix} 1&0&0\\ 0&-2&-1\\ 0&2&-1 \end{pmatrix}$

$\begin{pmatrix} 1&0&0\\ 0&0&1\\ 0&-2&-1 \end{pmatrix}^3=\begin{pmatrix} 1&0&0\\ 0&0&1\\ 0&-2&-1 \end{pmatrix}^2\begin{pmatrix} 1&0&0\\ 0&0&1\\ 0&-2&-1 \end{pmatrix}=$

$=\begin{pmatrix} 1&0&0\\ 0&-2&-1\\ 0&2&-1 \end{pmatrix}\begin{pmatrix} 1&0&0\\ 0&0&1\\ 0&-2&-1 \end{pmatrix}=\begin{pmatrix} 1&0&0\\ 0&2&-1\\ 0&2&3 \end{pmatrix}$

$-\begin{pmatrix} 1&0&0\\ 0&0&1\\ 0&-2&-1 \end{pmatrix}^3+\begin{pmatrix} 2&0&0\\ 0&2&0\\ 0&0&2 \end{pmatrix}=-\begin{pmatrix} 1&0&0\\ 0&2&-1\\ 0&2&3 \end{pmatrix}+\begin{pmatrix} 2&0&0\\ 0&2&0\\ 0&0&2 \end{pmatrix}=\begin{pmatrix} 1&0&0\\ 0&0&1\\ 0&-2&-1 \end{pmatrix}$

We obtain $f(T)\begin{pmatrix} x_1\\ x_2\\ x_3 \end{pmatrix}=\begin{pmatrix} 1&0&0\\ 0&0&1\\ 0&-2&-1 \end{pmatrix}\begin{pmatrix} x_1\\ x_2\\ x_3 \end{pmatrix}=T\begin{pmatrix} x_1\\ x_2\\ x_3 \end{pmatrix}$ , that is $f(T)=T$

Answer: $f(T)=T$