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Official PYQ · 2017

GATE CS 2017 Set 2 Q52 — Eigenvalues and Eigenvectors, LU Decomposition

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If the characteristic polynomial of a  3×33 \times 3  matrix MM over R\mathbb{R} (the set of real numbers) is λ3–4λ2+aλ+30,a∈R\lambda^3 – 4 \lambda^2 + a \lambda +30, \quad a \in \mathbb{R}, and one eigenvalue of MM is 2,2, then the largest among the absolute values of the eigenvalues of MM is _______
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Answer: 5

Let the characteristic polynomial of the 3×33 \times 3 matrix MM be P(λ)=λ3−4λ2+aλ+30P(\lambda) = \lambda^3 - 4\lambda^2 + a\lambda + 30.

Since λ=2\lambda = 2 is an eigenvalue of MM, it must be a root of P(λ)P(\lambda). Substituting λ=2\lambda = 2 into P(λ)=0P(\lambda) = 0 gives:

23−4(2)2+a(2)+30=02^3 - 4(2)^2 + a(2) + 30 = 08−16+2a+30=0  ⟹  2a+22=0  ⟹  a=−118 - 16 + 2a + 30 = 0 \implies 2a + 22 = 0 \implies a = -11

Let the three eigenvalues of MM be λ1,λ2,\lambda_1, \lambda_2, and λ3\lambda_3, with λ1=2\lambda_1 = 2. Using the properties of the roots of a polynomial:

  • Sum of eigenvalues: λ1+λ2+λ3=4  ⟹  2+λ2+λ3=4  ⟹  λ2+λ3=2\lambda_1 + \lambda_2 + \lambda_3 = 4 \implies 2 + \lambda_2 + \lambda_3 = 4 \implies \lambda_2 + \lambda_3 = 2
  • Product of eigenvalues: λ1λ2λ3=−30  ⟹  2λ2λ3=−30  ⟹  λ2λ3=−15\lambda_1 \lambda_2 \lambda_3 = -30 \implies 2 \lambda_2 \lambda_3 = -30 \implies \lambda_2 \lambda_3 = -15

Solving the quadratic equation x2−2x−15=0x^2 - 2x - 15 = 0 yields the other two eigenvalues:

(x−5)(x+3)=0  ⟹  x=5 or x=−3(x - 5)(x + 3) = 0 \implies x = 5 \text{ or } x = -3

Thus, the eigenvalues of MM are 2,5,2, 5, and −3-3. Their absolute values are ∣2∣=2|2| = 2, ∣5∣=5|5| = 5, and ∣−3∣=3|-3| = 3. The largest absolute value is 5\mathbf{5}.

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