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Official PYQ · 2015Set 3 · Q15Eigenvalues and Eigenvectors, LU DecompositionMCQ1 mark

GATE CS 2015 Set 3 Q15 — Eigenvalues and Eigenvectors, LU Decomposition

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In the given matrix [1−12010121]\begin{bmatrix} 1 & -1 & 2 \\ 0 & 1 & 0 \\ 1 & 2 & 1 \end{bmatrix} , one of the eigenvalues is 1.1. The eigenvectors corresponding to the eigenvalue 11 are

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Answer: B

To find the eigenvectors corresponding to the eigenvalue λ=1\lambda = 1 for the matrix A=[1−12010121]A = \begin{bmatrix} 1 & -1 & 2 \\ 0 & 1 & 0 \\ 1 & 2 & 1 \end{bmatrix}, we solve the characteristic equation (A−λI)X=0(A - \lambda I)X = 0 for λ=1\lambda = 1.

Substituting λ=1\lambda = 1 gives:

(A−I)X=[0−12000120][x1x2x3]=[000](A - I)X = \begin{bmatrix} 0 & -1 & 2 \\ 0 & 0 & 0 \\ 1 & 2 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}

This yields the following system of linear equations:

  1. −x2+2x3=0  ⟹  x2=2x3-x_2 + 2x_3 = 0 \implies x_2 = 2x_3
  2. x1+2x2=0  ⟹  x1=−2x2x_1 + 2x_2 = 0 \implies x_1 = -2x_2

Substituting x2=2x3x_2 = 2x_3 into the second equation:

x1=−2(2x3)=−4x3x_1 = -2(2x_3) = -4x_3

Expressing the eigenvector in terms of x3=ax_3 = a (where a≠0a \neq 0):

X=[x1x2x3]=[−4a2aa]=a[−421]X = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -4a \\ 2a \\ a \end{bmatrix} = a \begin{bmatrix} -4 \\ 2 \\ 1 \end{bmatrix}

Thus, the set of eigenvectors is {a(−4,2,1)∣a≠0,a∈R}\{a(-4, 2, 1) \mid a \neq 0, a \in \mathbb{R}\}, which corresponds to Option B.

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