GATE 2007 MA Eigen Value

Question

Comments

$(A)?$

---

A because eigen values are $0,-6,6$

$6A$ will have eigen values $6 \lambda $

---

@Mk Utkarsh

I'm not getting how you can say $(A)$ is the correct$?$

I know if $A$ having eigenvalue $\lambda$ then $6A$ having eigenvalue $6\lambda$

after that i don't know?

Answer 1

Let suppose $A=\begin{bmatrix} a_{11}&a_{12} &a_{13} \\ a_{21} &a_{22} &a_{23} \\a_{31} &a_{32} &a_{33} \end{bmatrix}_{3\times 3}$

Given that eigen values of $A$ is $-1,0,1$ with respective eigen vaector are $\begin{bmatrix} 1 &-1 &0 \end{bmatrix}^{T}=\begin{bmatrix} 1\\-1 \\0 \end{bmatrix},\begin{bmatrix} 1 &1 &-2 \end{bmatrix}^{T}=\begin{bmatrix} 1\\1 \\-2 \end{bmatrix},\begin{bmatrix} 1 &1 &1 \end{bmatrix}^{T}=\begin{bmatrix} 1\\1 \\1 \end{bmatrix}$

We know that $AX=\lambda X$

Lets take $\lambda =-1$ and it's corresponding eigen vector $X=\begin{bmatrix} 1\\-1 \\0 \end{bmatrix}$

$\begin{bmatrix} a_{11}&a_{12} &a_{13} \\ a_{21} &a_{22} &a_{23} \\a_{31} &a_{32} &a_{33} \end{bmatrix}_{3\times 3}.\begin{bmatrix} 1\\-1 \\0 \end{bmatrix}_{3\times 1}=-1.\begin{bmatrix} 1\\-1 \\0 \end{bmatrix}$

$\begin{bmatrix} a_{11}-a_{12}\\a_{21}-a_{22} \\a_{31}-a_{32} \end{bmatrix}=\begin{bmatrix} -1\\1 \\0 \end{bmatrix}$

Now we got $a_{11}-a_{12}=-1,a_{21}-a_{22}=1,a_{31}-a_{32}=0$

we can write like this $a_{11}-a_{12}=-1$

                                      $a_{21}-a_{22}=1$

                                      $a_{31}=a_{32}$

similarly for $\lambda=0$ it's corresponding eigen vector $X=\begin{bmatrix} 1\\1 \\-2 \end{bmatrix}$

and we got    $a_{11}+a_{12}-2a_{13}=0$

                         $a_{21}+a_{22}-2a_{23}=0$

                           $a_{31}+a_{32}-2a_{33}=0$

and similarly for $\lambda=0$ it's corresponding eigen vector $X=\begin{bmatrix} 1\\1 \\-2 \end{bmatrix}$

and we got    $a_{11}+a_{12}+a_{13}=1$

                         $a_{21}+a_{22}+a_{23}=1$

                           $a_{31}+a_{32}+a_{33}=1$

Try to solve it and get   $a_{11}=\frac{-1}{6},$  $a_{12}=\frac{5}{6},$  $a_{13}=\frac{1}{3}$

                                      $a_{21}=\frac{5}{6},$ $a_{22}=\frac{-1}{6},$ $a_{23}=\frac{1}{3}$

                                      $a_{31}=\frac{1}{3},$  $a_{32}=\frac{1}{3},$  $a_{33}=\frac{1}{3}$

Now $A=\begin{bmatrix} a_{11}&a_{12} &a_{13} \\ a_{21} &a_{22} &a_{23} \\a_{31} &a_{32} &a_{33} \end{bmatrix}=\begin{bmatrix} \frac{-1}{6}&\frac{5}{6} &\frac{1}{3}\\\frac{5}{6}&\frac{-1}{6} &\frac{1}{3} \\\frac{1}{3} &\frac{1}{3} &\frac{1}{3}\end{bmatrix}$

then $6A=\begin{bmatrix} {-1}&{5} &{2}\\ {5}&{-1}&{2} \\ {2} &{2} &{2}\end{bmatrix}$

So,$(A)$ is the correct answer.