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GATE 2024 EC – Question 55

Engineering Mathematics · Linear Algebra · 2 marks · Multiple select

Consider the matrix $\begin{bmatrix}1&k\\2&1\end{bmatrix}$, where $k$ is a positive real number. Which of the following vectors is/are eigenvector(s) of this matrix?

  1. $\begin{bmatrix}1\\-\sqrt{2/k}\end{bmatrix}$
  2. $\begin{bmatrix}1\\\sqrt{2/k}\end{bmatrix}$
  3. $\begin{bmatrix}\sqrt{2k}\\1\end{bmatrix}$
  4. $\begin{bmatrix}\sqrt{2k}\\-1\end{bmatrix}$

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Correct answer: (A) $\begin{bmatrix}1\\-\sqrt{2/k}\end{bmatrix}$; (B) $\begin{bmatrix}1\\\sqrt{2/k}\end{bmatrix}$

Explanation

The characteristic equation is $(1-\lambda)^2-2k=0$, so $\lambda=1\pm\sqrt{2k}$. For $\lambda=1+\sqrt{2k}$: $-\sqrt{2k}\,v_1+k\,v_2=0$ gives $v_2=\sqrt{2/k}\,v_1$, i.e. $[1,\ \sqrt{2/k}]^T$ (B). For $\lambda=1-\sqrt{2k}$: $v_2=-\sqrt{2/k}\,v_1$, i.e. $[1,\ -\sqrt{2/k}]^T$ (A). Checking (C): the first row gives $-\sqrt{2k}\sqrt{2k}+k=-k\ne0$ and (D) gives $2k-k=k\ne0$, so neither is an eigenvector.