# [SOLVED] cannot find 3rd eigen vector of this matrix for diagonalization

• Apr 28th 2010, 02:51 PM
thekrown
[SOLVED] cannot find 3rd eigen vector of this matrix for diagonalization
matrix A

2 0 0
0 2 0
1 1 3

eigen value #1 = 2
eigen vector #1 is

-1
1
0

eigen value #2 = 2
eigen vector #2 is

-1
0
1

eigen value #3 = 3
eigen vector #3 is

I cannot find this because I get this matrix after substituting llamda into the original matrix along the diagonal

-1 0 0
0 -1 0
1 1 0

• Apr 28th 2010, 03:02 PM
dwsmith
Quote:

Originally Posted by thekrown
matrix A

2 0 0
0 2 0
1 1 3

eigen value #1 = 2
eigen vector #1 is

-1
1
0

eigen value #2 = 2
eigen vector #2 is

-1
0
1

eigen value #3 = 3
eigen vector #3 is

I cannot find this because I get this matrix after substituting llamda into the original matrix along the diagonal

-1 0 0
0 -1 0
1 1 0

$\displaystyle \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0 \end{bmatrix}\Rightarrow x_3\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$
• Apr 28th 2010, 03:18 PM
thekrown
I don't understand how you end up with that. What must I do?
• Apr 28th 2010, 03:22 PM
dwsmith
$\displaystyle \begin{bmatrix} -1 & 0 & 0\\ 0 & -1 & 0\\ 1 & 1 & 0 \end{bmatrix}$ use elementary row operations to obtain the rref.

Multiple rows 1 and 2 by -1

$\displaystyle \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 1 & 1 & 0 \end{bmatrix}$

Subtract row 1 and 3 and row 2 and 3

$\displaystyle \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0 \end{bmatrix}$

$\displaystyle x_1=0$, $\displaystyle x_2=0$, and $\displaystyle x_3$ is a free variable.

$\displaystyle \begin{bmatrix} 0\\ 0\\ x_3 \end{bmatrix}\rightarrow x_3\begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}$
• Apr 28th 2010, 03:34 PM
thekrown
Okay if you don't mind I will say what I understood in my own words.

First I find the eigen values, subtract it from the diagonal. This gave us our matrix

1 0 0
0 1 0
1 1 0

from this point I can manupulate this matrix to obtain a reduced echelon form matrix.

Once I have my leading 1's, I can then find the value for x1, x2, x3.

In our case, it was simple once the reduction was made and we have the matrix

1 0 0
0 1 0
0 0 0

the third row can be removed, leaving us with a free variable for x3 and x1=0, x2=0.
• Apr 28th 2010, 03:36 PM
dwsmith
Quote:

Originally Posted by thekrown
Okay if you don't mind I will say what I understood in my own words.

First I find the eigen values, subtract it from the diagonal. This gave us our matrix

1 0 0
0 1 0
1 1 0

from this point I can manupulate this matrix to obtain a reduced echelon form matrix.

Once I have my leading 1's, I can then find the value for x1, x2, x3.

In our case, it was simple once the reduction was made and we have the matrix

1 0 0
0 1 0
0 0 0

the third row can be removed, leaving us with a free variable for x3 and x1=0, x2=0.

I wouldn't say the 3rd row can be removed. The third element can be any value.
• Apr 28th 2010, 03:42 PM
thekrown
Okay so I will leave it there, my teacher would remove it, maybe he was doing that to save chalk. Who knows.