# Eigen value , eigen vector

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• Jan 4th 2010, 07:52 PM
flintstone
Eigen value , eigen vector
Find the eigen value and eigen vector of $\begin{bmatrix} 1 & 2 & 3 \\ 3 & 1 & 0 \\ -2 & 0 & 1 \end{bmatrix}$

Solution :

$|A- \lambda I |$ = $\begin{bmatrix} (1 - \lambda) & 2 & 3 \\ 3 & (1 - \lambda) & 0 \\ -2 & 0 & (1 - \lambda) \end{bmatrix}$ =

$(1-\lambda)[(1 - \lambda)(1-\lambda) - 0]-2[3(1 - \lambda)-0]+3[0-(-2)(1 - \lambda)]$
= $(1-\lambda)(1 - 2 \lambda + 2)- 6 + 6 \lambda+ 6 - 6 \lambda$
$= 1 -2\lambda+\lambda^2-\lambda+2\lambda^2-\lambda^3$
$=(1-\lambda)(1-2\lambda+\lambda^2)$
$(1-\lambda)(\lambda-1)(\lambda-1)$

Eigen value $\lambda =1$

putting $\lambda=1$ in the matrix eq

$\begin{bmatrix} 0 & 2 & 3 \\ 3 & 0 & 0 \\ -2 & 0 & 0 \end{bmatrix}$ . $\begin{bmatrix}x \\ y \\z \end{bmatrix}$= $\begin{bmatrix} 0 \\ 0 \\0 \end{bmatrix}$ $\Rightarrow$ $\begin{cases}2y + 3z = 0 \\ 3x = 0 \\ -2x = 0 \end{cases}$

Stuck here !!!!!!!
• Jan 4th 2010, 09:20 PM
Shanks
$x=0,y=c,z=\frac{-2c}{3}$
the solution set is a subspace generated by $(0,1,\frac{-2}{3})$.
• Jan 5th 2010, 12:04 AM
flintstone
thanks
• Jan 5th 2010, 04:00 AM
HallsofIvy
Quote:

Originally Posted by flintstone
Find the eigen value and eigen vector of $\begin{bmatrix} 1 & 2 & 3 \\ 3 & 1 & 0 \\ -2 & 0 & 1 \end{bmatrix}$

Solution :

$|A- \lambda I |$ = $\begin{bmatrix} (1 - \lambda) & 2 & 3 \\ 3 & (1 - \lambda) & 0 \\ -2 & 0 & (1 - \lambda) \end{bmatrix}$ =

$(1-\lambda)[(1 - \lambda)(1-\lambda) - 0]-2[3(1 - \lambda)-0]+3[0-(-2)(1 - \lambda)]$
= $(1-\lambda)(1 - 2 \lambda + 2)- 6 + 6 \lambda+ 6 - 6 \lambda$
$= 1 -2\lambda+\lambda^2-\lambda+2\lambda^2-\lambda^3$
$=(1-\lambda)(1-2\lambda+\lambda^2)$
$(1-\lambda)(\lambda-1)(\lambda-1)$

Eigen value $\lambda =1$

putting $\lambda=1$ in the matrix eq

$\begin{bmatrix} 0 & 2 & 3 \\ 3 & 0 & 0 \\ -2 & 0 & 0 \end{bmatrix}$ . $\begin{bmatrix}x \\ y \\z \end{bmatrix}$= $\begin{bmatrix} 0 \\ 0 \\0 \end{bmatrix}$ $\Rightarrow$ $\begin{cases}2y + 3z = 0 \\ 3x = 0 \\ -2x = 0 \end{cases}$

Stuck here !!!!!!!

Both second and third equations just tell you that x= 0. The first equation tells you that 2y= -3z so y= -(3/2)z. Any eigenvector is of the form <0, -(3/2)z, z>= z<0, -3/2, 1>. If z= 2, that is <0, -3 , 2> so the space of eigenvectors of this matrix is spanned by that single vector (which is, of course, a multiple of the one Shanks gave).