# Basis for eigenspace!

• May 18th 2008, 06:47 AM
matty888
Basis for eigenspace!
Find a basis for the eigenspace corresponding to the eigenvalue λ=2 of the matrix

• May 18th 2008, 07:00 AM
Isomorphism
Quote:

Originally Posted by matty888
Find a basis for the eigenspace corresponding to the eigenvalue λ=2 of the matrix

Well assume a vector $[x, y, z]^{T}$ to be the eigenvector corresponding. By definition the eigenspace is the span of the eigenvector.
So
$|A - 2I| = 0 \Rightarrow 12x + y - 9z=0$

So the if x = a and y= b, we have $(a,b,\frac{12a+b}9)$

So $(1,0,\frac43)$ and $(0,1,\frac19)$ form the basis :)
• May 18th 2008, 07:06 AM
Moo
Hello,

Quote:

Originally Posted by matty888
Find a basis for the eigenspace corresponding to the eigenvalue λ=2 of the matrix

$A=\begin{pmatrix} 26 & 2 & -18 \\ 0 & 2 & 0 \\ 36 & 3 & -25 \end{pmatrix}$

$X_2=\begin{pmatrix} x \\ y \\ z \end{pmatrix}$

Solve for $X_2$ in $AX_2=2X_2$

This yields the following system :

$\left\{ \begin{array}{ccccccc} 26x &+ &2y&-&18z&=&2x \\ & & 2y & & & =& 2y \\ 36x&+&3y&-&25z&=&2z \end{array} \right.$

$\implies \left\{\begin{array}{ccccccc} 24x&+ &2y&-&18z&=&0 \\ 36x&+&3y&-&27z&=&0 \end{array} \right.$

(we can see that these two equations are actually the same).

There are two possibilities. Either y=1, either y=0.

If $y=0$, then $24x-18z=0 \implies x=\frac 34 z$. For example, $z=4$ and $x=3$

So $\begin{pmatrix} 3 \\ 0 \\ 4 \end{pmatrix}$ is an eigenvector of the eigenspace.

If $y=1$, then $24x+2-18z=0 \implies x=-\frac{1}{12}+\frac 34 z$. For example, $z=\frac 13$ and $x=\frac 16$

So $\begin{pmatrix} \frac 16 \\ 1 \\ \frac 13 \end{pmatrix}$ is the second eigenvector of the eigenspace.

Does it look clear ?
• May 18th 2008, 07:52 AM
matty888
Thanks(????)
Thanks guys just wondering are both of your answers legitimate,I am getting (3,0,4) and (-1,12,0) also, so not really sure????
• May 18th 2008, 08:11 AM
Isomorphism
Quote:

Originally Posted by matty888
Thanks guys just wondering are both of your answers legitimate,I am getting (3,0,4) and (-1,12,0) also, so not really sure????

Your answer is right too because from my basis I can represent

$(3,0,4) = 3(1,0,\frac43)$

$(-1,12,0) = -(1,0,\frac43) + 12(0,1,\frac19)$

So both of your vectors are a linear combination of my basis and your vectors need both of my basis vectors.So your set is a basis :D

You can try getting your vectors as a linear combination of Moo's basis (Nod)
• May 18th 2008, 08:14 AM
Moo
For my basis, it'll be :

$12(\frac 16,1,\frac 13)-(3,0,4)=(2,12,4)-(3,0,4)=(-1,12,0)$

(Wink)