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GATE 2016 EE – Question 38

Engineering Mathematics · Linear Algebra: Eigen values, Eigen vectors · 2 marks · Multiple choice

Let the eigenvalues of a $2 \times 2$ matrix $A$ be 1, $-2$ with eigenvectors $x_1$ and $x_2$ respectively. Then the eigenvalues and eigenvectors of the matrix $A^2 - 3A + 4I$ would, respectively, be

  1. 2, 14; $x_1$, $x_2$
  2. 2, 14; $x_1 + x_2$, $x_1 - x_2$
  3. 2, 0; $x_1$, $x_2$
  4. 2, 0; $x_1 + x_2$, $x_1 - x_2$

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Correct answer: (A) 2, 14; $x_1$, $x_2$

Explanation

A polynomial in $A$ has the same eigenvectors as $A$, and each eigenvalue $\lambda$ becomes $\lambda^2 - 3\lambda + 4$. For $\lambda = 1$ this is $1 - 3 + 4 = 2$, and for $\lambda = -2$ it is $4 + 6 + 4 = 14$. The eigenvectors stay $x_1$ and $x_2$.