1. ## Basis for eigenspace!

Find a basis for the eigenspace corresponding to the eigenvalue λ=2 of the matrix

A =

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

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

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

So $\displaystyle (1,0,\frac43)$ and $\displaystyle (0,1,\frac19)$ form the basis

3. Hello,

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

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

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

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

This yields the following system :

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

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

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

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

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

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

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

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

Does it look clear ?

4. ## 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????

5. 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

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

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

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

You can try getting your vectors as a linear combination of Moo's basis

6. For my basis, it'll be :

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