# Thread: Finding Eigen Value and Eigen Vectors

1. ## Finding Eigen Value and Eigen Vectors

Find the Eigen value and Eigen Vectors of the matrix
$\displaystyle \begin{bmatrix} 1 & -6 & -4 \\ 0 & 4 & 2 \\ 0 & -6 & -3 \\ \end{bmatrix}$

2. Originally Posted by varunnayudu
Find the Eigen value and Eigen Vectors of the matrix
$\displaystyle \begin{matrix} 1 & -6 & -4 \\ 0 & 4 & 2 \\ 0 & -6 & -3 \\ \end{matrix}$
Where is the trouble here? There are two eigenvalues: $\displaystyle \lambda = 0$ and $\displaystyle \lambda = 1$ (repeated).

Getting the eigenvectors does not look difficult.

eg. A possible eigenvector (other forms are possible) asociated with $\displaystyle \lambda = 0$ is $\displaystyle \left(\begin{array}{c} -2 \\ 1 \\ -2 \end{array}\right)$, found by solving

$\displaystyle x - 6y - 4z = 0$ .... (1)

$\displaystyle 4y + 2z = 0 \Rightarrow 2y + z = 0$ .... (2)

$\displaystyle -6y - 3z = 0 \Rightarrow 2y + z = 0$ .... (3)

Where are you stuck?

3. ## Is this right?

$\displaystyle \Rightarrow$$\displaystyle \begin{bmatrix} 1-\lambda & -6 & -4 \\ 0 & 4- \lambda & 2 \\ 0 & -6 & -3 - \lambda \end{bmatrix} \displaystyle (1- \lambda){(4- \lambda)(-3- \lambda)+12} + 0 +0 \displaystyle (1- \lambda)(- \lambda + \lambda ^2) \displaystyle \lambda = 0,1,1 for \displaystyle \lambda = 0 weget \displaystyle \begin{bmatrix} 1 & -6 & -4 \\ 0 & 4 & 2 \\ 0 & -6 & -3 \end{bmatrix} \displaystyle \begin{bmatrix} x \\ y \\ z \end{bmatrix} =\displaystyle \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} or\displaystyle \begin{bmatrix} x-6y-4z=0 \\ 0x +4y+2z=0 \end{bmatrix} \displaystyle \frac{x}{-12+16} = \frac{y}{2-0} = \frac{z}{4-0} \displaystyle \frac{x}{4} = \frac{y}{2} = \frac{z}{4} Can u tell me hw u got 2 -1 2 instead of 2 1 2 4. Originally Posted by varunnayudu \displaystyle \Rightarrow$$\displaystyle \begin{bmatrix} 1-\lambda & -6 & -4 \\ 0 & 4- \lambda & 2 \\ 0 & -6 & -3 - \lambda \end{bmatrix}$
$\displaystyle (1- \lambda){(4- \lambda)(-3- \lambda)+12} + 0 +0$
$\displaystyle (1- \lambda)(- \lambda + \lambda ^2)$
$\displaystyle \lambda = 0,1,1$

for
$\displaystyle \lambda = 0$ weget
$\displaystyle \begin{bmatrix} 1 & -6 & -4 \\ 0 & 4 & 2 \\ 0 & -6 & -3 \end{bmatrix}$ $\displaystyle \begin{bmatrix} x \\ y \\ z \end{bmatrix}$=$\displaystyle \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

or$\displaystyle \begin{bmatrix} x-6y-4z=0 \\ 0x +4y+2z=0 \end{bmatrix}$

$\displaystyle \frac{x}{-12+16} = \frac{y}{2-0} = \frac{z}{4-0}$

$\displaystyle \frac{x}{4} = \frac{y}{2} = \frac{z}{4}$

Can u tell me hw u got 2 -1 2 instead of 2 1 2
x - 6y - 4z = 0 .... (1)

2y + z = 0 .... (2)

Let y = t. Then:

From (2): z = -2y = -2t.

From (1): x = 6y + 4z = 6t - 8t = -2t.

Other forms of the solution are possible.