Question

The eigen values and the corresponding eigen vectors of a $2\times 2$ matrix are given by
Eigen value Eigen vector
$I_1=8$ $v_1=\left[\begin{array}{c}1\\ 1\end{array}\right]$
$I_2=8$ $v_2=\left[\begin{array}{c}1\\ -1\end{array}\right]$
The matrix is

• A. $\left[\begin{array}{cc}6& 2\\ 2& 6\end{array}\right]$
• B. $\left[\begin{array}{cc}4& 6\\ 6& 4\end{array}\right]$
• C. $\left[\begin{array}{cc}2& 4\\ 4& 2\end{array}\right]$
• D. $\left[\begin{array}{cc}4& 8\\ 8& 4\end{array}\right]$

Answer 1

The correct option is A $\left[\begin{array}{cc}6& 2\\ 2& 6\end{array}\right]$

Let $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$
$Ax=\lambda x$
$\Rightarrow \left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{c}1\\ 1\end{array}\right]=8\left[\begin{array}{c}1\\ 1\end{array}\right]$
$\Rightarrow a+b=8$ & $c+d=8$ .... (1)
$\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{c}1\\ -1\end{array}\right]=4\left[\begin{array}{c}1\\ -1\end{array}\right]$
$a-b=4$ & $c-d=-4$ .... (2)
Solving (1) and (2)
$a=6,b=2,c=2,d=6$
$A=\left[\begin{array}{cc}6& 2\\ 2& 6\end{array}\right]$
Alternative solution:
Sum of eigen values = trace (A) is satisfied by option (a)