Question

The pair of eigen vectors corresponding to the two eigen values of the matrix $\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$ is

• A. $\left[\begin{array}{c}1\\ -j\end{array}\right],\left[\begin{array}{c}j\\ -1\end{array}\right]$
• B. $\left[\begin{array}{c}0\\ 1\end{array}\right],\left[\begin{array}{c}-1\\ 0\end{array}\right]$
• C. $\left[\begin{array}{c}1\\ j\end{array}\right],\left[\begin{array}{c}0\\ 1\end{array}\right]$
• D. $\left[\begin{array}{c}1\\ j\end{array}\right],\left[\begin{array}{c}j\\ 1\end{array}\right]$

Answer 1

The correct option is A $\left[\begin{array}{c}1\\ -j\end{array}\right],\left[\begin{array}{c}j\\ -1\end{array}\right]$

Given $A=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$ so C.Equation is $|A=\lambda I|=0$
$\Rightarrow \left|\begin{array}{cc}0-\lambda & -1\\ 1& 0-\lambda \end{array}\right|=0\Rightarrow \lambda =\pm j$

Now consider $AX=\lambda X$
$\Rightarrow$ $(A-\lambda I)X=0$

or $\left[\begin{array}{cc}0-\lambda & 1\\ 1& 0-\lambda \end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$ ..(1)

Case-I : Put $\lambda =j$ in equation (1)

i.e. $\left[\begin{array}{cc}-j& -1\\ 1& -j\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$
$-j{x}_1-x_2=0\Rightarrow x_1=j{x}_2$

so $X_1=\left[\begin{array}{c}jk\\ k\end{array}\right]\sim \left[\begin{array}{c}j\\ 1\end{array}\right]$
Similarly when we put $\lambda =-j$ in (1)
We get $X_2=\left[\begin{array}{c}-j\\ 1\end{array}\right]$

So $X_1=\left[\begin{array}{c}j\\ 1\end{array}\right]\stackrel{\times (-j)}{\to }\left[\begin{array}{c}1\\ -j\end{array}\right]$

and $X_2=\left[\begin{array}{c}-j\\ 1\end{array}\right]\stackrel{\times (-1)}{\to }\left[\begin{array}{c}j\\ -1\end{array}\right]$
So option (a.)