← 2024 Paper 1
UPSC 2024 Maths Optional Paper 1 Q4a — Step-by-Step Solution 20 marks · Section A
Eigenvalues and eigenvectors · Linear Algebra · asked 10× in 14 yrs · Read the full method →
Question
Let A = ( 3 2 4 2 0 2 4 2 3 ) A=\begin{pmatrix}3 & 2 & 4\\2 & 0 & 2\\4 & 2 & 3\end{pmatrix} A = 3 2 4 2 0 2 4 2 3 A A A A − 15 A^{-15} A − 15 A − 15 = ( A − 1 ) 15 A^{-15}=(A^{-1})^{15} A − 15 = ( A − 1 ) 15
Technique
Characteristic polynomial via 3 × 3 3\times3 3 × 3 ( A − λ I ) v = 0 (A-\lambda I)v=0 ( A − λ I ) v = 0 A − 15 A^{-15} A − 15 A A A λ − 15 \lambda^{-15} λ − 15
Solution
Eigenvalues of A A A
Characteristic equation det ( A − λ I ) = 0 \det(A-\lambda I)=0 det ( A − λ I ) = 0
det ( 3 − λ 2 4 2 − λ 2 4 2 3 − λ ) = 0. \det\begin{pmatrix}3-\lambda & 2 & 4\\2 & -\lambda & 2\\4 & 2 & 3-\lambda\end{pmatrix}=0. det 3 − λ 2 4 2 − λ 2 4 2 3 − λ = 0. Expanding and collecting: λ 3 − 6 λ 2 − 15 λ − 8 = 0 \lambda^3-6\lambda^2-15\lambda-8=0 λ 3 − 6 λ 2 − 15 λ − 8 = 0
Testing λ = − 1 \lambda=-1 λ = − 1 − 1 − 6 + 15 − 8 = 0 -1-6+15-8=0 − 1 − 6 + 15 − 8 = 0 ( λ + 1 ) (\lambda+1) ( λ + 1 )
( λ + 1 ) ( λ 2 − 7 λ − 8 ) = ( λ + 1 ) ( λ + 1 ) ( λ − 8 ) = ( λ + 1 ) 2 ( λ − 8 ) . (\lambda+1)(\lambda^2-7\lambda-8)=(\lambda+1)(\lambda+1)(\lambda-8)=(\lambda+1)^2(\lambda-8). ( λ + 1 ) ( λ 2 − 7 λ − 8 ) = ( λ + 1 ) ( λ + 1 ) ( λ − 8 ) = ( λ + 1 ) 2 ( λ − 8 ) . Eigenvalues: λ = 8 \lambda=8 λ = 8 λ = − 1 \lambda=-1 λ = − 1
Sanity check. Trace: 3 + 0 + 3 = 6 = 8 + ( − 1 ) + ( − 1 ) 3+0+3=6=8+(-1)+(-1) 3 + 0 + 3 = 6 = 8 + ( − 1 ) + ( − 1 ) 8 = 8 ⋅ ( − 1 ) ⋅ ( − 1 ) 8=8\cdot(-1)\cdot(-1) 8 = 8 ⋅ ( − 1 ) ⋅ ( − 1 )
Eigenvectors of A A A
For λ = 8 \lambda=8 λ = 8 solve ( A − 8 I ) v = 0 (A-8I)v=0 ( A − 8 I ) v = 0
( − 5 2 4 2 − 8 2 4 2 − 5 ) → v 1 = 2 v 2 , v 3 = 2 v 2 . \begin{pmatrix}-5 & 2 & 4\\2 & -8 & 2\\4 & 2 & -5\end{pmatrix}\;\to\; v_1=2v_2,\;v_3=2v_2. − 5 2 4 2 − 8 2 4 2 − 5 → v 1 = 2 v 2 , v 3 = 2 v 2 . Setting v 2 = 1 v_2=1 v 2 = 1 v ( 8 ) = ( 2 , 1 , 2 ) T v^{(8)}=(2,1,2)^T v ( 8 ) = ( 2 , 1 , 2 ) T
For λ = − 1 \lambda=-1 λ = − 1 solve ( A + I ) v = 0 (A+I)v=0 ( A + I ) v = 0
( 4 2 4 2 1 2 4 2 4 ) . \begin{pmatrix}4 & 2 & 4\\2 & 1 & 2\\4 & 2 & 4\end{pmatrix}. 4 2 4 2 1 2 4 2 4 . All rows reduce to 2 v 1 + v 2 + 2 v 3 = 0 2v_1+v_2+2v_3=0 2 v 1 + v 2 + 2 v 3 = 0
v a ( − 1 ) = ( 1 , − 2 , 0 ) T , v b ( − 1 ) = ( 0 , − 2 , 1 ) T . v^{(-1)}_a=(1,-2,0)^T,\qquad v^{(-1)}_b=(0,-2,1)^T. v a ( − 1 ) = ( 1 , − 2 , 0 ) T , v b ( − 1 ) = ( 0 , − 2 , 1 ) T . Eigenvalues and eigenvectors of A − 15 A^{-15} A − 15
Since A A A det A = 8 ≠ 0 \det A=8\ne 0 det A = 8 = 0 A − 1 A^{-1} A − 1 A v = λ v Av=\lambda v A v = λ v A − k v = λ − k v A^{-k}v=\lambda^{-k}v A − k v = λ − k v A A A
λ \lambda λ A A A λ − 15 \lambda^{-15} λ − 15 A − 15 A^{-15} A − 15 Eigenvector(s) 8 8 8 8 − 15 = 2 − 45 8^{-15}=2^{-45} 8 − 15 = 2 − 45 ( 2 , 1 , 2 ) T (2,1,2)^T ( 2 , 1 , 2 ) T − 1 -1 − 1 ( − 1 ) − 15 = − 1 (-1)^{-15}=-1 ( − 1 ) − 15 = − 1 ( 1 , − 2 , 0 ) T (1,-2,0)^T ( 1 , − 2 , 0 ) T ( 0 , − 2 , 1 ) T (0,-2,1)^T ( 0 , − 2 , 1 ) T
Answer
Eigenvalues of A : 8 (eigenvector ( 2 , 1 , 2 ) T ) , − 1 multiplicity 2 (basis ( 1 , − 2 , 0 ) T , ( 0 , − 2 , 1 ) T ) . \boxed{\;\text{Eigenvalues of }A:\;8\text{ (eigenvector }(2,1,2)^T),\;-1\text{ multiplicity 2 (basis }(1,-2,0)^T,(0,-2,1)^T).\;} Eigenvalues of A : 8 (eigenvector ( 2 , 1 , 2 ) T ) , − 1 multiplicity 2 (basis ( 1 , − 2 , 0 ) T , ( 0 , − 2 , 1 ) T ) . Eigenvalues of A − 15 : 2 − 45 and − 1 (multiplicity 2), same eigenvectors as A. \boxed{\;\text{Eigenvalues of }A^{-15}:\;2^{-45}\text{ and }-1\text{ (multiplicity 2),\text{ same eigenvectors as }A.\;}\;} Eigenvalues of A − 15 : 2 − 45 and − 1 (multiplicity 2), same eigenvectors as A.