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

**thekrown** · April 28th 2010, 02:51 PM

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

I am stuck here. Please help.

**dwsmith** · April 28th 2010, 03:02 PM

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

I am stuck here. Please help.

$\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0 \end{bmatrix}\Rightarrow x_3\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$

**thekrown** · April 28th 2010, 03:18 PM

I don't understand how you end up with that. What must I do?

**dwsmith** · April 28th 2010, 03:22 PM

$\begin{bmatrix} -1 & 0 & 0\\ 0 & -1 & 0\\ 1 & 1 & 0 \end{bmatrix}$ use elementary row operations to obtain the rref.

Multiple rows 1 and 2 by -1

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

Subtract row 1 and 3 and row 2 and 3

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

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

$\begin{bmatrix} 0\\ 0\\ x_3 \end{bmatrix}\rightarrow x_3\begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}$

**thekrown** · April 28th 2010, 03:34 PM

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.

**dwsmith** · April 28th 2010, 03:36 PM

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.

**thekrown** · April 28th 2010, 03:42 PM

Okay so I will leave it there, my teacher would remove it, maybe he was doing that to save chalk. Who knows.

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