GATE 2024 EC Q.55: solution and answer

Official answer: A, BSource: Official answer key, GATE 2024 EC (IISc Bengaluru)

1. Call the matrix $M=\left[\begin{array}{cc}1& k\\ 2& 1\end{array}\right]$ (so it is not confused with option (A)). Characteristic equation: $\det (M-\lambda I)=(1-\lambda {)}^2-2k=0$, so $\lambda -1=\pm \sqrt{2k}$, i.e. ${\lambda }_{1,2}=1\pm \sqrt{2k}$. Because $k>0$ these are real and distinct. (Check: their sum is $2$, the trace.)

2. Eigenvector condition. Take the first row of $(M-\lambda I)\left[\begin{array}{c}x\\ y\end{array}\right]=0$: $$(1-\lambda )x+ky=0\Rightarrow y=\frac{\lambda -1}{k}x$$

3. For $\lambda =1+\sqrt{2k}$: $y=\frac{\sqrt{2k}}{k}x=\sqrt{\frac{2k}{k^2}}x=\sqrt{2/k}x$. With $x=1$ the eigenvector is $\left[\begin{array}{c}1\\ \sqrt{2/k}\end{array}\right]$, which is option (B).

4. For $\lambda =1-\sqrt{2k}$: $y=-\frac{\sqrt{2k}}{k}x=-\sqrt{2/k}x$. With $x=1$ the eigenvector is $\left[\begin{array}{c}1\\ -\sqrt{2/k}\end{array}\right]$, which is option (A).

5. Direct check of (A): $$M\left[\begin{array}{c}1\\ -\sqrt{2/k}\end{array}\right]=\left[\begin{array}{c}1-k\sqrt{2/k}\\ 2-\sqrt{2/k}\end{array}\right]=\left[\begin{array}{c}1-\sqrt{2k}\\ 2-\sqrt{2/k}\end{array}\right]$$ and $(1-\sqrt{2k})\left(-\sqrt{2/k}\right)=-\sqrt{2/k}+\sqrt{4}=2-\sqrt{2/k}$. So the result is $(1-\sqrt{2k})\left[\begin{array}{c}1\\ -\sqrt{2/k}\end{array}\right]$, confirming (A) belongs to $\lambda =1-\sqrt{2k}$.

6. Options (C) and (D) have $y/x=\pm \frac{1}{\sqrt{2k}}$, but every eigenvector of $M$ has $y/x=\pm \sqrt{2/k}$. Squaring, equality would need $\frac{1}{2k}=\frac{2}{k}$, i.e. $1=4$, which is false for every $k>0$. So neither (C) nor (D) is an eigenvector. For instance, $M\left[\begin{array}{c}\sqrt{2k}\\ 1\end{array}\right]=\left[\begin{array}{c}\sqrt{2k}+k\\ 2\sqrt{2k}+1\end{array}\right]$, which is not a multiple of $\left[\begin{array}{c}\sqrt{2k}\\ 1\end{array}\right]$.

7. Correct options: (A) and (B), one eigenvector for each of the two distinct eigenvalues.