Question

An eigen vector of $P=\left[\begin{array}{ccc}1& 1& 0\\ 0& 2& 2\\ 0& 0& 3\end{array}\right]$ is

• A. ${\left[\begin{array}{ccc}-1& 1& 1\end{array}\right]}^T$
• B. ${\left[\begin{array}{ccc}1& 2& 1\end{array}\right]}^T$
• C. ${\left[\begin{array}{ccc}1& -1& 2\end{array}\right]}^T$
• D. ${\left[\begin{array}{ccc}2& 1& -1\end{array}\right]}^T$

Answer 1

The correct option is B ${\left[\begin{array}{ccc}1& 2& 1\end{array}\right]}^T$

$P=\left[\begin{array}{ccc}1& 1& 0\\ 0& 2& 2\\ 0& 0& 3\end{array}\right]$

We know that "The eigen values upper triangular matrix (UTM) or lower triangular matrix (LTM) are the leading diagonal elements"
$\therefore$ Eigen values of matrix P are 1, 2 and 3
Let ${\lambda }_1=1,{\lambda }_2=2,{\lambda }_3=3$
Eigen vector for $\lambda =3$
$[P-\lambda I]\left[X\right]=0$

$\left[\begin{array}{ccc}1-3& 1& 0\\ 0& 2-3& 2\\ 0& 0& 3-3\end{array}\right]\left[\begin{array}{c}{X}_1\\ X_2\\ X_3\end{array}\right]=0$
$\Rightarrow \left[\begin{array}{ccc}-2& 1& 0\\ 0& -1& 2\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}{X}_1\\ X_2\\ X_3\end{array}\right]=0$
$\Rightarrow 2x_1=x_2,x_2=2x_3$
Now let $x_3=k$. Then $x_2=2k$ & $x_1=k$
$\Rightarrow \left[\begin{array}{ccc}{x}_1:& x_2:& x_3\end{array}\right]=\left[\begin{array}{ccc}1& :2& :1\end{array}\right]$
$\therefore$ Eigen vector corresponding to eigen value 3 is ${\left[\begin{array}{ccc}1:& 2:& 1\end{array}\right]}^T$