# Finding Eigen Value and Eigen Vectors

• December 23rd 2008, 12:54 AM
varunnayudu
Finding Eigen Value and Eigen Vectors
Find the Eigen value and Eigen Vectors of the matrix
$
\begin{bmatrix}
1 & -6 & -4 \\
0 & 4 & 2 \\
0 & -6 & -3 \\

\end{bmatrix}
$
• December 23rd 2008, 01:47 AM
mr fantastic
Quote:

Originally Posted by varunnayudu
Find the Eigen value and Eigen Vectors of the matrix
$
\begin{matrix}
1 & -6 & -4 \\
0 & 4 & 2 \\
0 & -6 & -3 \\
\end{matrix}
$

Where is the trouble here? There are two eigenvalues: $\lambda = 0$ and $\lambda = 1$ (repeated).

Getting the eigenvectors does not look difficult.

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

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

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

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

Where are you stuck?
• December 26th 2008, 04:30 AM
varunnayudu
Is this right?
http://www.mathhelpforum.com/math-he...9df91814-1.gif

$\Rightarrow$ $\begin{bmatrix}
1-\lambda & -6 & -4 \\
0 & 4- \lambda & 2 \\
0 & -6 & -3 - \lambda
\end{bmatrix}$

$(1- \lambda){(4- \lambda)(-3- \lambda)+12} + 0 +0$
$(1- \lambda)(- \lambda + \lambda ^2)$
$\lambda = 0,1,1$

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

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

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

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

Can u tell me hw u got 2 -1 2 instead of 2 1 2
• December 26th 2008, 04:43 AM
mr fantastic
Quote:

Originally Posted by varunnayudu
http://www.mathhelpforum.com/math-he...9df91814-1.gif

$\Rightarrow$ $\begin{bmatrix}
1-\lambda & -6 & -4 \\
0 & 4- \lambda & 2 \\
0 & -6 & -3 - \lambda
\end{bmatrix}$

$(1- \lambda){(4- \lambda)(-3- \lambda)+12} + 0 +0$
$(1- \lambda)(- \lambda + \lambda ^2)$
$\lambda = 0,1,1$

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

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

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

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

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.