Answer to Question #144441 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 that $T:\R^3\rightarrow \R^3$ is linear operator given by

$$T(x_1,x_2,x_3)=(x_1,x_3,-2x_2-x_3)$$

Consider the standard basis of $\R^3$ which are $e_1,e_2,e_3$ , thus

$$T(e_1)=a_1e_1+a_2e_2+a_3e_3=(a_1,a_2,a_3)=(1,0,0)\\ T(e_2)=b_1e_1+b_2e_2+b_3e_3=(b_1,b_2,b_3)=(0,0,-2)\\ T(e_3)=c_1e_1+c_2e_2+c_3e_3=(c_1,c_2,c_3)=(0,1,-1)$$

Thus, matrix of T with respect to standard basis is

$$M=M(T)_v^v=\begin{bmatrix} 1 &0&0\\ 0&0&-2\\ 0&1&-1 \end{bmatrix}$$

Where $v=(e_1,e_2,e_3)$ .

Now, polynomial of T is given $f(x)=-x^3+2$ , thus operator

$$f(T)=-T^3+3I\iff -M^3+2I_{3\times 3}$$

Now,

$$M^3=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 2 & 2 \\ 0 & -1 & 3 \end{array} \right)\\ \implies -M^3=\left( \begin{array}{ccc} -1 & 0 & 0 \\ 0 &- 2 & -2 \\ 0 & 1 & -3 \end{array} \right)\\ \implies -M^3+2I_{3\times 3}=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0& -2 \\ 0 & 1 & -1\end{array} \right)\\$$

Thus,

$$f(T)=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0& -2 \\ 0 & 1 & -1\end{array} \right)\\$$