# a little trouble finding the eigen vector here...

• Apr 28th 2010, 04:05 AM
thekrown
a little trouble finding the eigen vector here...
So I have original matrix

2 2 -2
0 1 0
0 0 1

with eigen values 2, 1, 1

i can find the eigen vector for the llamda 1 which is

2
0
1

and

-2
1
0

but I cannot find the eigen vector for value 2

I get this matrix

0 2 -2
0 -1 0
0 0 -1

• Apr 28th 2010, 04:26 AM
HallsofIvy
Directly from the definition of "eigenvector": $\displaystyle Av= \lambda v$, if $\displaystyle \begin{bmatrix} x \\ y \\ z\end{bmatrix}$ is an eigenvector of A corresponding to eigenvalue 2, then
$\displaystyle \begin{bmatrix}2 & 2 & -2 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= 2\begin{bmatrix} x \\ y \\ z\end{bmatrix}$
$\displaystyle \begin{bmatrix}2x+ 2y- 2z \\ y \\ z\end{bmatrix}= \begin{bmatrix}2x \\ 2y \\ 2z\end{bmatrix}$
so we have the three equations 2x+ 2y- 2z= 2x, y= 2y, z= 2z. The last two give y= 0, z= 0, of course, and 2x= 2x+ 0- 0 is satisfied for any x.

That is, of course, equivalent to saying $\displaystyle \begin{bmatrix}0 & 2 & -2 \\ 0 & -1 & 0 \\ 0 & 0 & -1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}$
which gives the equations 2y- 2z= 0, y= 0, z= 0. That says nothing about x so x can be any number.
• Apr 28th 2010, 05:26 AM
thekrown
So the eigen vector could be

1
0
0

when constructing the matrix made up of the eigen vectors but in columns, is there an order I must respect?

The way I would go about it is to use llamda 1 (eigen value 1) for the first column, second for second and third for third. I get this matrix

1 2 -2
0 0 1
0 1 0

I think I can swap the 2nd and thrd column since their share the same eigen value. Is that true?