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
Hello,
$\displaystyle A=\begin{pmatrix} 26 & 2 & -18 \\ 0 & 2 & 0 \\ 36 & 3 & -25 \end{pmatrix}$
$\displaystyle X_2=\begin{pmatrix} x \\ y \\ z \end{pmatrix}$
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 ?
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