← 2026 Paper 1
UPSC 2026 Maths Optional Paper 1 Q2a — Step-by-Step Solution 20 marks · Section A
Eigenvalues and eigenvectors · Linear Algebra · asked 10× in 14 yrs · Read the full method →
Question
Let T : P 2 [ x ] → P 2 [ x ] T : P_2[x] \to P_2[x] T : P 2 [ x ] → P 2 [ x ]
T ( a 0 + a 1 x + a 2 x 2 ) = − ( a 0 + 2 a 1 + a 2 ) + ( 2 a 0 + 3 a 1 ) x − 2 ( a 0 + a 1 ) x 2 T(a_0 + a_1 x + a_2 x^2) = -(a_0 + 2a_1 + a_2) + (2a_0 + 3a_1)x - 2(a_0 + a_1)x^2 T ( a 0 + a 1 x + a 2 x 2 ) = − ( a 0 + 2 a 1 + a 2 ) + ( 2 a 0 + 3 a 1 ) x − 2 ( a 0 + a 1 ) x 2 where P 2 [ x ] P_2[x] P 2 [ x ] x x x ≤ 2 \leq 2 ≤ 2 a 0 , a 1 , a 2 ∈ R a_0, a_1, a_2 \in \mathbb{R} a 0 , a 1 , a 2 ∈ R T T T T T T P 2 [ x ] P_2[x] P 2 [ x ]
Technique
Build [ T ] B [T]_B [ T ] B T ( 1 ) , T ( x ) , T ( x 2 ) T(1), T(x), T(x^2) T ( 1 ) , T ( x ) , T ( x 2 ) B = { 1 , x , x 2 } B=\{1,x,x^2\} B = { 1 , x , x 2 } det ( A − λ I ) = 0 \det(A-\lambda I)=0 det ( A − λ I ) = 0 A − λ I A-\lambda I A − λ I translate each coordinate triple back into a polynomial of P 2 [ x ] P_2[x] P 2 [ x ] T T T P 2 [ x ] P_2[x] P 2 [ x ] R 3 \mathbb{R}^3 R 3
Solution
Step 1 — Fix the basis and the coordinate convention.
Take the standard (ordered) basis B = { 1 , x , x 2 } B=\{1,\;x,\;x^2\} B = { 1 , x , x 2 } P 2 [ x ] P_2[x] P 2 [ x ] dim P 2 [ x ] = 3 \dim P_2[x]=3 dim P 2 [ x ] = 3 p = a 0 + a 1 x + a 2 x 2 p = a_0+a_1x+a_2x^2 p = a 0 + a 1 x + a 2 x 2
[ p ] B = ( a 0 a 1 a 2 ) . [p]_B = \begin{pmatrix}a_0\\a_1\\a_2\end{pmatrix}. [ p ] B = a 0 a 1 a 2 . The matrix A = [ T ] B A=[T]_B A = [ T ] B [ T ( p ) ] B = A [ p ] B [T(p)]_B = A\,[p]_B [ T ( p ) ] B = A [ p ] B the j j j A A A j j j (Writing the images as rows is the standard slip here; it produces A T A^{\mathsf T} A T
Step 2 — Compute the images of the basis vectors.
Read off the rule T ( a 0 + a 1 x + a 2 x 2 ) = − ( a 0 + 2 a 1 + a 2 ) + ( 2 a 0 + 3 a 1 ) x − 2 ( a 0 + a 1 ) x 2 T(a_0+a_1x+a_2x^2) = -(a_0+2a_1+a_2) + (2a_0+3a_1)x - 2(a_0+a_1)x^2 T ( a 0 + a 1 x + a 2 x 2 ) = − ( a 0 + 2 a 1 + a 2 ) + ( 2 a 0 + 3 a 1 ) x − 2 ( a 0 + a 1 ) x 2
For 1 1 1 a 0 = 1 , a 1 = 0 , a 2 = 0 a_0=1,\,a_1=0,\,a_2=0 a 0 = 1 , a 1 = 0 , a 2 = 0
T ( 1 ) = − ( 1 + 0 + 0 ) + ( 2 + 0 ) x − 2 ( 1 + 0 ) x 2 = − 1 + 2 x − 2 x 2 , [ T ( 1 ) ] B = ( − 1 2 − 2 ) . T(1) = -(1+0+0) + (2+0)x - 2(1+0)x^2 = -1 + 2x - 2x^2,\qquad [T(1)]_B=\begin{pmatrix}-1\\2\\-2\end{pmatrix}. T ( 1 ) = − ( 1 + 0 + 0 ) + ( 2 + 0 ) x − 2 ( 1 + 0 ) x 2 = − 1 + 2 x − 2 x 2 , [ T ( 1 ) ] B = − 1 2 − 2 . For x x x a 0 = 0 , a 1 = 1 , a 2 = 0 a_0=0,\,a_1=1,\,a_2=0 a 0 = 0 , a 1 = 1 , a 2 = 0
T ( x ) = − ( 0 + 2 + 0 ) + ( 0 + 3 ) x − 2 ( 0 + 1 ) x 2 = − 2 + 3 x − 2 x 2 , [ T ( x ) ] B = ( − 2 3 − 2 ) . T(x) = -(0+2+0) + (0+3)x - 2(0+1)x^2 = -2 + 3x - 2x^2,\qquad [T(x)]_B=\begin{pmatrix}-2\\3\\-2\end{pmatrix}. T ( x ) = − ( 0 + 2 + 0 ) + ( 0 + 3 ) x − 2 ( 0 + 1 ) x 2 = − 2 + 3 x − 2 x 2 , [ T ( x ) ] B = − 2 3 − 2 . For x 2 x^2 x 2 a 0 = 0 , a 1 = 0 , a 2 = 1 a_0=0,\,a_1=0,\,a_2=1 a 0 = 0 , a 1 = 0 , a 2 = 1
T ( x 2 ) = − ( 0 + 0 + 1 ) + 0 ⋅ x − 2 ( 0 + 0 ) x 2 = − 1 , [ T ( x 2 ) ] B = ( − 1 0 0 ) . T(x^2) = -(0+0+1) + 0\cdot x - 2(0+0)x^2 = -1,\qquad [T(x^2)]_B=\begin{pmatrix}-1\\0\\0\end{pmatrix}. T ( x 2 ) = − ( 0 + 0 + 1 ) + 0 ⋅ x − 2 ( 0 + 0 ) x 2 = − 1 , [ T ( x 2 ) ] B = − 1 0 0 . Step 3 — Assemble the matrix representation.
Stacking these three coordinate vectors as columns,
A = [ T ] B = ( − 1 − 2 − 1 2 3 0 − 2 − 2 0 ) \boxed{\;A=[T]_B=\begin{pmatrix}-1 & -2 & -1\\ 2 & 3 & 0\\ -2 & -2 & 0\end{pmatrix}\;} A = [ T ] B = − 1 2 − 2 − 2 3 − 2 − 1 0 0 Check the construction against the defining rule:
A ( a 0 a 1 a 2 ) = ( − a 0 − 2 a 1 − a 2 2 a 0 + 3 a 1 − 2 a 0 − 2 a 1 ) , A\begin{pmatrix}a_0\\a_1\\a_2\end{pmatrix}=\begin{pmatrix}-a_0-2a_1-a_2\\ 2a_0+3a_1\\ -2a_0-2a_1\end{pmatrix}, A a 0 a 1 a 2 = − a 0 − 2 a 1 − a 2 2 a 0 + 3 a 1 − 2 a 0 − 2 a 1 , which is exactly ( − ( a 0 + 2 a 1 + a 2 ) , 2 a 0 + 3 a 1 , − 2 ( a 0 + a 1 ) ) T \big(-(a_0+2a_1+a_2),\;2a_0+3a_1,\;-2(a_0+a_1)\big)^{\mathsf T} ( − ( a 0 + 2 a 1 + a 2 ) , 2 a 0 + 3 a 1 , − 2 ( a 0 + a 1 ) ) T
Step 4 — Characteristic polynomial.
det ( A − λ I ) = ∣ − 1 − λ − 2 − 1 2 3 − λ 0 − 2 − 2 − λ ∣ . \det(A-\lambda I)=\begin{vmatrix}-1-\lambda & -2 & -1\\ 2 & 3-\lambda & 0\\ -2 & -2 & -\lambda\end{vmatrix}. det ( A − λ I ) = − 1 − λ 2 − 2 − 2 3 − λ − 2 − 1 0 − λ . Expand along the third row (it carries a zero-free but simple pattern):
( 3 , 1 ) (3,1) ( 3 , 1 ) − 2 -2 − 2 ( − 1 ) 3 + 1 = + 1 (-1)^{3+1}=+1 ( − 1 ) 3 + 1 = + 1 ∣ − 2 − 1 3 − λ 0 ∣ = 0 − ( − 1 ) ( 3 − λ ) = 3 − λ \begin{vmatrix}-2 & -1\\ 3-\lambda & 0\end{vmatrix}=0-(-1)(3-\lambda)=3-\lambda − 2 3 − λ − 1 0 = 0 − ( − 1 ) ( 3 − λ ) = 3 − λ − 2 ( 3 − λ ) = 2 λ − 6 -2(3-\lambda)=2\lambda-6 − 2 ( 3 − λ ) = 2 λ − 6 ( 3 , 2 ) (3,2) ( 3 , 2 ) − 2 -2 − 2 ( − 1 ) 3 + 2 = − 1 (-1)^{3+2}=-1 ( − 1 ) 3 + 2 = − 1 ∣ − 1 − λ − 1 2 0 ∣ = 0 + 2 = 2 \begin{vmatrix}-1-\lambda & -1\\ 2 & 0\end{vmatrix}=0+2=2 − 1 − λ 2 − 1 0 = 0 + 2 = 2 ( − 2 ) ( − 1 ) ( 2 ) = 4 (-2)(-1)(2)=4 ( − 2 ) ( − 1 ) ( 2 ) = 4 ( 3 , 3 ) (3,3) ( 3 , 3 ) − λ -\lambda − λ + 1 +1 + 1 ∣ − 1 − λ − 2 2 3 − λ ∣ = ( − 1 − λ ) ( 3 − λ ) + 4 = λ 2 − 2 λ − 3 + 4 = λ 2 − 2 λ + 1 \begin{vmatrix}-1-\lambda & -2\\ 2 & 3-\lambda\end{vmatrix}=(-1-\lambda)(3-\lambda)+4=\lambda^2-2\lambda-3+4=\lambda^2-2\lambda+1 − 1 − λ 2 − 2 3 − λ = ( − 1 − λ ) ( 3 − λ ) + 4 = λ 2 − 2 λ − 3 + 4 = λ 2 − 2 λ + 1 − λ ( λ 2 − 2 λ + 1 ) -\lambda(\lambda^2-2\lambda+1) − λ ( λ 2 − 2 λ + 1 )
Summing,
det ( A − λ I ) = − λ 3 + 2 λ 2 − λ + 2 λ − 6 + 4 = − λ 3 + 2 λ 2 + λ − 2 = − ( λ 3 − 2 λ 2 − λ + 2 ) . \det(A-\lambda I) = -\lambda^3+2\lambda^2-\lambda+2\lambda-6+4 = -\lambda^3+2\lambda^2+\lambda-2 = -\big(\lambda^3-2\lambda^2-\lambda+2\big). det ( A − λ I ) = − λ 3 + 2 λ 2 − λ + 2 λ − 6 + 4 = − λ 3 + 2 λ 2 + λ − 2 = − ( λ 3 − 2 λ 2 − λ + 2 ) . Factor by grouping:
λ 3 − 2 λ 2 − λ + 2 = λ 2 ( λ − 2 ) − ( λ − 2 ) = ( λ − 2 ) ( λ 2 − 1 ) = ( λ − 2 ) ( λ − 1 ) ( λ + 1 ) . \lambda^3-2\lambda^2-\lambda+2=\lambda^2(\lambda-2)-(\lambda-2)=(\lambda-2)(\lambda^2-1)=(\lambda-2)(\lambda-1)(\lambda+1). λ 3 − 2 λ 2 − λ + 2 = λ 2 ( λ − 2 ) − ( λ − 2 ) = ( λ − 2 ) ( λ 2 − 1 ) = ( λ − 2 ) ( λ − 1 ) ( λ + 1 ) . So the characteristic equation is ( λ − 2 ) ( λ − 1 ) ( λ + 1 ) = 0 (\lambda-2)(\lambda-1)(\lambda+1)=0 ( λ − 2 ) ( λ − 1 ) ( λ + 1 ) = 0
λ = 1 , − 1 , 2. \lambda = 1,\;-1,\;2. λ = 1 , − 1 , 2. Consistency check: tr A = − 1 + 3 + 0 = 2 = 1 + ( − 1 ) + 2 \operatorname{tr}A=-1+3+0=2=1+(-1)+2 tr A = − 1 + 3 + 0 = 2 = 1 + ( − 1 ) + 2 det A = − 2 = ( 1 ) ( − 1 ) ( 2 ) \det A = -2 = (1)(-1)(2) det A = − 2 = ( 1 ) ( − 1 ) ( 2 )
Step 5 — Eigenvector for λ 1 = 1 \lambda_1=1 λ 1 = 1
( A − I ) v = 0 (A-I)v=0 ( A − I ) v = 0
A − I = ( − 2 − 2 − 1 2 2 0 − 2 − 2 − 1 ) . A-I=\begin{pmatrix}-2 & -2 & -1\\ 2 & 2 & 0\\ -2 & -2 & -1\end{pmatrix}. A − I = − 2 2 − 2 − 2 2 − 2 − 1 0 − 1 . Row 2 gives 2 a 0 + 2 a 1 = 0 ⇒ a 1 = − a 0 2a_0+2a_1=0\Rightarrow a_1=-a_0 2 a 0 + 2 a 1 = 0 ⇒ a 1 = − a 0 − 2 a 0 − 2 a 1 − a 2 = 0 ⇒ − 2 a 0 + 2 a 0 − a 2 = 0 ⇒ a 2 = 0 -2a_0-2a_1-a_2=0\Rightarrow -2a_0+2a_0-a_2=0\Rightarrow a_2=0 − 2 a 0 − 2 a 1 − a 2 = 0 ⇒ − 2 a 0 + 2 a 0 − a 2 = 0 ⇒ a 2 = 0 a 0 = 1 a_0=1 a 0 = 1
v 1 = ( 1 , − 1 , 0 ) T ⟺ p 1 ( x ) = 1 − x . v_1=(1,-1,0)^{\mathsf T}\quad\Longleftrightarrow\quad p_1(x)=1-x. v 1 = ( 1 , − 1 , 0 ) T ⟺ p 1 ( x ) = 1 − x . Direct check in P 2 [ x ] P_2[x] P 2 [ x ] with a 0 = 1 , a 1 = − 1 , a 2 = 0 a_0=1,a_1=-1,a_2=0 a 0 = 1 , a 1 = − 1 , a 2 = 0 T ( 1 − x ) = − ( 1 − 2 + 0 ) + ( 2 − 3 ) x − 2 ( 1 − 1 ) x 2 = 1 − x = 1 ⋅ p 1 T(1-x)=-(1-2+0)+(2-3)x-2(1-1)x^2 = 1-x = 1\cdot p_1 T ( 1 − x ) = − ( 1 − 2 + 0 ) + ( 2 − 3 ) x − 2 ( 1 − 1 ) x 2 = 1 − x = 1 ⋅ p 1
Step 6 — Eigenvector for λ 2 = − 1 \lambda_2=-1 λ 2 = − 1
( A + I ) v = 0 (A+I)v=0 ( A + I ) v = 0
A + I = ( 0 − 2 − 1 2 4 0 − 2 − 2 1 ) . A+I=\begin{pmatrix}0 & -2 & -1\\ 2 & 4 & 0\\ -2 & -2 & 1\end{pmatrix}. A + I = 0 2 − 2 − 2 4 − 2 − 1 0 1 . Row 1: − 2 a 1 − a 2 = 0 ⇒ a 2 = − 2 a 1 -2a_1-a_2=0\Rightarrow a_2=-2a_1 − 2 a 1 − a 2 = 0 ⇒ a 2 = − 2 a 1 2 a 0 + 4 a 1 = 0 ⇒ a 0 = − 2 a 1 2a_0+4a_1=0\Rightarrow a_0=-2a_1 2 a 0 + 4 a 1 = 0 ⇒ a 0 = − 2 a 1 − 2 ( − 2 a 1 ) − 2 a 1 + ( − 2 a 1 ) = 4 a 1 − 2 a 1 − 2 a 1 = 0 -2(-2a_1)-2a_1+(-2a_1)=4a_1-2a_1-2a_1=0 − 2 ( − 2 a 1 ) − 2 a 1 + ( − 2 a 1 ) = 4 a 1 − 2 a 1 − 2 a 1 = 0 a 1 = − 1 a_1=-1 a 1 = − 1
v 2 = ( 2 , − 1 , 2 ) T ⟺ p 2 ( x ) = 2 − x + 2 x 2 . v_2=(2,-1,2)^{\mathsf T}\quad\Longleftrightarrow\quad p_2(x)=2-x+2x^2. v 2 = ( 2 , − 1 , 2 ) T ⟺ p 2 ( x ) = 2 − x + 2 x 2 . Direct check: with a 0 = 2 , a 1 = − 1 , a 2 = 2 a_0=2,a_1=-1,a_2=2 a 0 = 2 , a 1 = − 1 , a 2 = 2 T ( p 2 ) = − ( 2 − 2 + 2 ) + ( 4 − 3 ) x − 2 ( 2 − 1 ) x 2 = − 2 + x − 2 x 2 = − ( 2 − x + 2 x 2 ) = ( − 1 ) p 2 T(p_2)=-(2-2+2)+(4-3)x-2(2-1)x^2 = -2+x-2x^2 = -(2-x+2x^2)=(-1)\,p_2 T ( p 2 ) = − ( 2 − 2 + 2 ) + ( 4 − 3 ) x − 2 ( 2 − 1 ) x 2 = − 2 + x − 2 x 2 = − ( 2 − x + 2 x 2 ) = ( − 1 ) p 2
Step 7 — Eigenvector for λ 3 = 2 \lambda_3=2 λ 3 = 2
( A − 2 I ) v = 0 (A-2I)v=0 ( A − 2 I ) v = 0
A − 2 I = ( − 3 − 2 − 1 2 1 0 − 2 − 2 − 2 ) . A-2I=\begin{pmatrix}-3 & -2 & -1\\ 2 & 1 & 0\\ -2 & -2 & -2\end{pmatrix}. A − 2 I = − 3 2 − 2 − 2 1 − 2 − 1 0 − 2 . Row 2: 2 a 0 + a 1 = 0 ⇒ a 1 = − 2 a 0 2a_0+a_1=0\Rightarrow a_1=-2a_0 2 a 0 + a 1 = 0 ⇒ a 1 = − 2 a 0 a 0 + a 1 + a 2 = 0 ⇒ a 2 = − a 0 − a 1 = − a 0 + 2 a 0 = a 0 a_0+a_1+a_2=0\Rightarrow a_2=-a_0-a_1=-a_0+2a_0=a_0 a 0 + a 1 + a 2 = 0 ⇒ a 2 = − a 0 − a 1 = − a 0 + 2 a 0 = a 0 − 3 a 0 − 2 ( − 2 a 0 ) − a 0 = − 3 a 0 + 4 a 0 − a 0 = 0 -3a_0-2(-2a_0)-a_0=-3a_0+4a_0-a_0=0 − 3 a 0 − 2 ( − 2 a 0 ) − a 0 = − 3 a 0 + 4 a 0 − a 0 = 0 a 0 = 1 a_0=1 a 0 = 1
v 3 = ( 1 , − 2 , 1 ) T ⟺ p 3 ( x ) = 1 − 2 x + x 2 = ( 1 − x ) 2 . v_3=(1,-2,1)^{\mathsf T}\quad\Longleftrightarrow\quad p_3(x)=1-2x+x^2=(1-x)^2. v 3 = ( 1 , − 2 , 1 ) T ⟺ p 3 ( x ) = 1 − 2 x + x 2 = ( 1 − x ) 2 . Direct check: with a 0 = 1 , a 1 = − 2 , a 2 = 1 a_0=1,a_1=-2,a_2=1 a 0 = 1 , a 1 = − 2 , a 2 = 1 T ( p 3 ) = − ( 1 − 4 + 1 ) + ( 2 − 6 ) x − 2 ( 1 − 2 ) x 2 = 2 − 4 x + 2 x 2 = 2 p 3 T(p_3)=-(1-4+1)+(2-6)x-2(1-2)x^2 = 2-4x+2x^2 = 2\,p_3 T ( p 3 ) = − ( 1 − 4 + 1 ) + ( 2 − 6 ) x − 2 ( 1 − 2 ) x 2 = 2 − 4 x + 2 x 2 = 2 p 3
Step 8 — Diagonalisability.
The three eigenvalues 1 , − 1 , 2 1,-1,2 1 , − 1 , 2 distinct , so each has algebraic multiplicity 1 1 1 1 1 1 { 1 − x , 2 − x + 2 x 2 , 1 − 2 x + x 2 } \{1-x,\;2-x+2x^2,\;1-2x+x^2\} { 1 − x , 2 − x + 2 x 2 , 1 − 2 x + x 2 } B ′ B' B ′ P 2 [ x ] P_2[x] P 2 [ x ] T T T
[ T ] B ′ = diag ( 1 , − 1 , 2 ) , P − 1 A P = diag ( 1 , − 1 , 2 ) , P = ( 1 2 1 − 1 − 1 − 2 0 2 1 ) . [T]_{B'}=\operatorname{diag}(1,-1,2),\qquad P^{-1}AP = \operatorname{diag}(1,-1,2),\quad P=\begin{pmatrix}1 & 2 & 1\\ -1 & -1 & -2\\ 0 & 2 & 1\end{pmatrix}. [ T ] B ′ = diag ( 1 , − 1 , 2 ) , P − 1 A P = diag ( 1 , − 1 , 2 ) , P = 1 − 1 0 2 − 1 2 1 − 2 1 . (No multiplicity is repeated, so no defect can arise.)
Answer
[ T ] B = ( − 1 − 2 − 1 2 3 0 − 2 − 2 0 ) ; λ = 1 : p 1 ( x ) = 1 − x λ = − 1 : p 2 ( x ) = 2 − x + 2 x 2 λ = 2 : p 3 ( x ) = 1 − 2 x + x 2 \boxed{\;[T]_B=\begin{pmatrix}-1 & -2 & -1\\ 2 & 3 & 0\\ -2 & -2 & 0\end{pmatrix};\qquad
\begin{aligned}
\lambda&=1: & p_1(x)&=1-x\\
\lambda&=-1: & p_2(x)&=2-x+2x^2\\
\lambda&=2: & p_3(x)&=1-2x+x^2
\end{aligned}\;} [ T ] B = − 1 2 − 2 − 2 3 − 2 − 1 0 0 ; λ λ λ = 1 : = − 1 : = 2 : p 1 ( x ) p 2 ( x ) p 3 ( x ) = 1 − x = 2 − x + 2 x 2 = 1 − 2 x + x 2 (eigenvectors up to nonzero scalar multiples; T T T