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UPSC 2016 Maths Optional Paper 1 Q2a-i — Step-by-Step Solution 10 marks · Section A
Matrix of a linear transformation · Linear Algebra · asked 10× in 14 yrs · Read the full method →
Question
If M 2 ( R ) M_2(\mathbb R) M 2 ( R ) 2 × 2 2\times2 2 × 2 P 2 ( x ) P_2(x) P 2 ( x ) T : M 2 ( R ) → P 2 ( x ) T:M_2(\mathbb R)\to P_2(x) T : M 2 ( R ) → P 2 ( x ) T ( [ a b c d ] ) = a + c + ( a − d ) x + ( b + c ) x 2 T\!\left(\begin{bmatrix}a & b\\ c & d\end{bmatrix}\right)=a+c+(a-d)x+(b+c)x^2 T ( [ a c b d ] ) = a + c + ( a − d ) x + ( b + c ) x 2 M 2 ( R ) M_2(\mathbb R) M 2 ( R ) P 2 ( x ) P_2(x) P 2 ( x ) T T T
Technique
Coordinates of T ( basis element ) T(\text{basis element}) T ( basis element )
Solution
Step 1 — Choose ordered bases
Domain basis (standard): E 11 = ( 1 0 0 0 ) , E 12 = ( 0 1 0 0 ) , E 21 = ( 0 0 1 0 ) , E 22 = ( 0 0 0 1 ) . E_{11}=\left(\begin{smallmatrix}1&0\\0&0\end{smallmatrix}\right),\ E_{12}=\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right),\ E_{21}=\left(\begin{smallmatrix}0&0\\1&0\end{smallmatrix}\right),\ E_{22}=\left(\begin{smallmatrix}0&0\\0&1\end{smallmatrix}\right). E 11 = ( 1 0 0 0 ) , E 12 = ( 0 0 1 0 ) , E 21 = ( 0 1 0 0 ) , E 22 = ( 0 0 0 1 ) . { 1 , x , x 2 } \{1,\,x,\,x^2\} { 1 , x , x 2 }
Step 2 — Images of the basis matrices
Using T ( a b c d ) = ( a + c ) + ( a − d ) x + ( b + c ) x 2 T\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)=(a+c)+(a-d)x+(b+c)x^2 T ( a c b d ) = ( a + c ) + ( a − d ) x + ( b + c ) x 2
T ( E 11 ) = 1 + x + 0 ⋅ x 2 → ( 1 , 1 , 0 ) ⊤ T ( E 12 ) = 0 + 0 ⋅ x + 1 ⋅ x 2 → ( 0 , 0 , 1 ) ⊤ T ( E 21 ) = 1 + 0 ⋅ x + 1 ⋅ x 2 → ( 1 , 0 , 1 ) ⊤ T ( E 22 ) = 0 − x + 0 ⋅ x 2 → ( 0 , − 1 , 0 ) ⊤ . \begin{aligned}T(E_{11})&=1+x+0\cdot x^2 &&\to (1,1,0)^{\!\top}\\ T(E_{12})&=0+0\cdot x+1\cdot x^2 &&\to (0,0,1)^{\!\top}\\ T(E_{21})&=1+0\cdot x+1\cdot x^2 &&\to (1,0,1)^{\!\top}\\ T(E_{22})&=0-x+0\cdot x^2 &&\to (0,-1,0)^{\!\top}.\end{aligned} T ( E 11 ) T ( E 12 ) T ( E 21 ) T ( E 22 ) = 1 + x + 0 ⋅ x 2 = 0 + 0 ⋅ x + 1 ⋅ x 2 = 1 + 0 ⋅ x + 1 ⋅ x 2 = 0 − x + 0 ⋅ x 2 → ( 1 , 1 , 0 ) ⊤ → ( 0 , 0 , 1 ) ⊤ → ( 1 , 0 , 1 ) ⊤ → ( 0 , − 1 , 0 ) ⊤ . Step 3 — Assemble the matrix (images as columns)
[ T ] = [ 1 0 1 0 1 0 0 − 1 0 1 1 0 ] ( 3 × 4 ) . \boxed{\;[T]=\begin{bmatrix}1&0&1&0\\ 1&0&0&-1\\ 0&1&1&0\end{bmatrix}\quad(3\times4).\;} [ T ] = 1 1 0 0 0 1 1 0 1 0 − 1 0 ( 3 × 4 ) . Step 4 — Null space of T T T
Solve T ( a b c d ) = 0 T\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)=0 T ( a c b d ) = 0
a + c = 0 , a − d = 0 , b + c = 0. a+c=0,\qquad a-d=0,\qquad b+c=0. a + c = 0 , a − d = 0 , b + c = 0. From these: c = − a c=-a c = − a d = a d=a d = a b = − c = a b=-c=a b = − c = a a = t a=t a = t
( a , b , c , d ) = t ( 1 , 1 , − 1 , 1 ) . (a,b,c,d)=t(1,1,-1,1). ( a , b , c , d ) = t ( 1 , 1 , − 1 , 1 ) . Answer
ker T = { t [ 1 1 − 1 1 ] : t ∈ R } , dim ( ker T ) = 1. \boxed{\;\ker T=\left\{\,t\begin{bmatrix}1&1\\-1&1\end{bmatrix}:t\in\mathbb R\right\},\qquad \dim(\ker T)=1.\;} ker T = { t [ 1 − 1 1 1 ] : t ∈ R } , dim ( ker T ) = 1.