Question

The matrix $M=\left[\begin{array}{ccc}-2& 2& -3\\ 2& 1& 6\\ -1& -2& 0\end{array}\right]$ has eigen values 3, -3, 5. An eigen vector corresponding to the eigen value 5 is $[12-1{]}^T$. One of the eigen vector of the matrix $M^3$ is

• A. ${\left[\begin{array}{ccc}1& 8& -1\end{array}\right]}^T$
• B. ${\left[\begin{array}{ccc}1& 2& -1\end{array}\right]}^T$
• C. ${\left[\begin{array}{ccc}1& \sqrt[3]{32}& -1\end{array}\right]}^T$
• D. ${\left[\begin{array}{ccc}1& 1& -1\end{array}\right]}^T$

Answer 1

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

Consider $AX=\lambda X$, where X is an eigen vector of A
$\begin{array}{ccc}{A}^{2}X& =& {\lambda }^{2}X\\ ⋮& & ⋮\\ ⋮& & ⋮\\ ⋮& & ⋮\\ A^{m}X& =& {\lambda }^{m}X\end{array}$

(where ${\lambda }^m$ is the Eigen value of $A^m$)
(where X is an Eigen vector of $A^m$)

$A^{m}X={\lambda }^{m}X$, where X is an eigen vector of $A^m$

Hence A and $A^m$ have same eigen vectors.

So $[12-1{]}^T$ is one of the eigen vector of $M^3$ also.