Question #144441
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).
1
Expert's answer
2020-11-18T19:07:37-0500

Given that T:R3R3T:\R^3\rightarrow \R^3 is linear operator given by


T(x1,x2,x3)=(x1,x3,2x2x3)T(x_1,x_2,x_3)=(x_1,x_3,-2x_2-x_3)

Consider the standard basis of R3\R^3 which are e1,e2,e3e_1,e_2,e_3 , thus


T(e1)=a1e1+a2e2+a3e3=(a1,a2,a3)=(1,0,0)T(e2)=b1e1+b2e2+b3e3=(b1,b2,b3)=(0,0,2)T(e3)=c1e1+c2e2+c3e3=(c1,c2,c3)=(0,1,1)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)vv=[100002011]M=M(T)_v^v=\begin{bmatrix} 1 &0&0\\ 0&0&-2\\ 0&1&-1 \end{bmatrix}

Where v=(e1,e2,e3)v=(e_1,e_2,e_3) .


Now, polynomial of T is given f(x)=x3+2f(x)=-x^3+2 , thus operator


f(T)=T3+3I    M3+2I3×3f(T)=-T^3+3I\iff -M^3+2I_{3\times 3}

Now,

M3=(100022013)    M3=(100022013)    M3+2I3×3=(100002011)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)=(100002011)f(T)=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0& -2 \\ 0 & 1 & -1\end{array} \right)\\


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS