← 2016 Paper 1
UPSC 2016 Maths Optional Paper 1 Q2c — Step-by-Step Solution 18 marks · Section A
Matrix of a linear transformation · Linear Algebra · asked 10× in 13 yrs · Read the full method →
Question
If A = [ 1 − 1 2 − 2 1 − 1 1 2 3 ] A=\begin{bmatrix}1 & -1 & 2\\ -2 & 1 & -1\\ 1 & 2 & 3\end{bmatrix} A = 1 − 2 1 − 1 1 2 2 − 1 3 T : P 2 ( x ) → P 2 ( x ) T:P_2(x)\to P_2(x) T : P 2 ( x ) → P 2 ( x ) { 1 − x , x ( 1 − x ) , x ( 1 + x ) } \{1-x,\ x(1-x),\ x(1+x)\} { 1 − x , x ( 1 − x ) , x ( 1 + x )} { 1 , 1 + x , 1 + x 2 } \{1,1+x,1+x^2\} { 1 , 1 + x , 1 + x 2 } T T T
Technique
Columns of A A A C C C T ( b j ) T(b_j) T ( b j ) T ( b j ) T(b_j) T ( b j ) B B B T T T
Solution
Domain (ordered) basis B = { b 1 = 1 − x , b 2 = x − x 2 , b 3 = x + x 2 } B=\{\,b_1=1-x,\ b_2=x-x^2,\ b_3=x+x^2\,\} B = { b 1 = 1 − x , b 2 = x − x 2 , b 3 = x + x 2 } C = { c 1 = 1 , c 2 = 1 + x , c 3 = 1 + x 2 } C=\{\,c_1=1,\ c_2=1+x,\ c_3=1+x^2\,\} C = { c 1 = 1 , c 2 = 1 + x , c 3 = 1 + x 2 } j j j A A A C C C T ( b j ) T(b_j) T ( b j )
Step 1 — Reconstruct T T T
T ( b j ) = ∑ i A i j c i T(b_j)=\sum_i A_{ij}\,c_i T ( b j ) = ∑ i A ij c i
T ( b 1 ) T(b_1) T ( b 1 ) ( 1 , − 2 , 1 ) (1,-2,1) ( 1 , − 2 , 1 ) 1 ⋅ 1 − 2 ( 1 + x ) + 1 ( 1 + x 2 ) = 1 − 2 − 2 x + 1 + x 2 = x 2 − 2 x . 1\cdot1-2(1+x)+1(1+x^2)=1-2-2x+1+x^2=x^2-2x. 1 ⋅ 1 − 2 ( 1 + x ) + 1 ( 1 + x 2 ) = 1 − 2 − 2 x + 1 + x 2 = x 2 − 2 x .
T ( b 2 ) T(b_2) T ( b 2 ) ( − 1 , 1 , 2 ) (-1,1,2) ( − 1 , 1 , 2 ) − 1 ⋅ 1 + 1 ( 1 + x ) + 2 ( 1 + x 2 ) = − 1 + 1 + x + 2 + 2 x 2 = 2 x 2 + x + 2. -1\cdot1+1(1+x)+2(1+x^2)=-1+1+x+2+2x^2=2x^2+x+2. − 1 ⋅ 1 + 1 ( 1 + x ) + 2 ( 1 + x 2 ) = − 1 + 1 + x + 2 + 2 x 2 = 2 x 2 + x + 2.
T ( b 3 ) T(b_3) T ( b 3 ) ( 2 , − 1 , 3 ) (2,-1,3) ( 2 , − 1 , 3 ) 2 ⋅ 1 − 1 ( 1 + x ) + 3 ( 1 + x 2 ) = 2 − 1 − x + 3 + 3 x 2 = 3 x 2 − x + 4. 2\cdot1-1(1+x)+3(1+x^2)=2-1-x+3+3x^2=3x^2-x+4. 2 ⋅ 1 − 1 ( 1 + x ) + 3 ( 1 + x 2 ) = 2 − 1 − x + 3 + 3 x 2 = 3 x 2 − x + 4.
So
T ( 1 − x ) = x 2 − 2 x , T ( x − x 2 ) = 2 x 2 + x + 2 , T ( x + x 2 ) = 3 x 2 − x + 4. T(1-x)=x^2-2x,\quad T(x-x^2)=2x^2+x+2,\quad T(x+x^2)=3x^2-x+4. T ( 1 − x ) = x 2 − 2 x , T ( x − x 2 ) = 2 x 2 + x + 2 , T ( x + x 2 ) = 3 x 2 − x + 4. Step 2 — Express 1 , x , x 2 1,\,x,\,x^2 1 , x , x 2 B B B
Solve for coordinates ( α , β , γ ) (\alpha,\beta,\gamma) ( α , β , γ ) α b 1 + β b 2 + γ b 3 = \alpha b_1+\beta b_2+\gamma b_3= α b 1 + β b 2 + γ b 3 =
1 = 1 ⋅ b 1 + 1 2 b 2 + 1 2 b 3 1=1\cdot b_1+\tfrac12 b_2+\tfrac12 b_3 1 = 1 ⋅ b 1 + 2 1 b 2 + 2 1 b 3 x = 0 ⋅ b 1 + 1 2 b 2 + 1 2 b 3 x=0\cdot b_1+\tfrac12 b_2+\tfrac12 b_3 x = 0 ⋅ b 1 + 2 1 b 2 + 2 1 b 3 x 2 = 0 ⋅ b 1 − 1 2 b 2 + 1 2 b 3 x^2=0\cdot b_1-\tfrac12 b_2+\tfrac12 b_3 x 2 = 0 ⋅ b 1 − 2 1 b 2 + 2 1 b 3
Step 3 — Apply linearity to get T ( 1 ) , T ( x ) , T ( x 2 ) T(1),T(x),T(x^2) T ( 1 ) , T ( x ) , T ( x 2 )
T ( 1 ) = T ( b 1 ) + 1 2 T ( b 2 ) + 1 2 T ( b 3 ) = ( x 2 − 2 x ) + 1 2 ( 2 x 2 + x + 2 ) + 1 2 ( 3 x 2 − x + 4 ) = 7 2 x 2 − 2 x + 3. T(1)=T(b_1)+\tfrac12 T(b_2)+\tfrac12 T(b_3)=(x^2-2x)+\tfrac12(2x^2+x+2)+\tfrac12(3x^2-x+4)=\tfrac72x^2-2x+3. T ( 1 ) = T ( b 1 ) + 2 1 T ( b 2 ) + 2 1 T ( b 3 ) = ( x 2 − 2 x ) + 2 1 ( 2 x 2 + x + 2 ) + 2 1 ( 3 x 2 − x + 4 ) = 2 7 x 2 − 2 x + 3. T ( x ) = 1 2 T ( b 2 ) + 1 2 T ( b 3 ) = 1 2 ( 2 x 2 + x + 2 ) + 1 2 ( 3 x 2 − x + 4 ) = 5 2 x 2 + 3. T(x)=\tfrac12 T(b_2)+\tfrac12 T(b_3)=\tfrac12(2x^2+x+2)+\tfrac12(3x^2-x+4)=\tfrac52x^2+3. T ( x ) = 2 1 T ( b 2 ) + 2 1 T ( b 3 ) = 2 1 ( 2 x 2 + x + 2 ) + 2 1 ( 3 x 2 − x + 4 ) = 2 5 x 2 + 3. T ( x 2 ) = − 1 2 T ( b 2 ) + 1 2 T ( b 3 ) = − 1 2 ( 2 x 2 + x + 2 ) + 1 2 ( 3 x 2 − x + 4 ) = 1 2 x 2 − x + 1. T(x^2)=-\tfrac12 T(b_2)+\tfrac12 T(b_3)=-\tfrac12(2x^2+x+2)+\tfrac12(3x^2-x+4)=\tfrac12x^2-x+1. T ( x 2 ) = − 2 1 T ( b 2 ) + 2 1 T ( b 3 ) = − 2 1 ( 2 x 2 + x + 2 ) + 2 1 ( 3 x 2 − x + 4 ) = 2 1 x 2 − x + 1. For f = p + q x + r x 2 f=p+qx+rx^2 f = p + q x + r x 2 T ( f ) = p T ( 1 ) + q T ( x ) + r T ( x 2 ) T(f)=p\,T(1)+q\,T(x)+r\,T(x^2) T ( f ) = p T ( 1 ) + q T ( x ) + r T ( x 2 )
Answer
T ( p + q x + r x 2 ) = ( 3 p + 3 q + r ) + ( − 2 p − r ) x + 1 2 ( 7 p + 5 q + r ) x 2 . \boxed{\;T(p+qx+rx^2)=(3p+3q+r)+(-2p-r)\,x+\tfrac12(7p+5q+r)\,x^2.\;} T ( p + q x + r x 2 ) = ( 3 p + 3 q + r ) + ( − 2 p − r ) x + 2 1 ( 7 p + 5 q + r ) x 2 .