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.
Quote:
Originally Posted by thekrown
I don't understand how you end up with that. What must I do?
use elementary row operations to obtain the rref.
Multiple rows 1 and 2 by -1
Subtract row 1 and 3 and row 2 and 3
,
, and
is a free variable.
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.Quote:
Originally Posted by thekrown
Okay so I will leave it there, my teacher would remove it, maybe he was doing that to save chalk. Who knows.