MHB Eigenvector of 3x3 matrix with complex eigenvalues

rayne1
Messages
32
Reaction score
0
Matrix A:
0 -6 10
-2 12 -20
-1 6 -10

I got the eigenvalues of: 0, 1+i, and 1-i. I can find the eigenvector of the eigenvalue 0, but for the complex eigenvalues, I keep on getting the reduced row echelon form of:
1 0 0 | 0
0 1 0 | 0
0 0 1 | 0

So, how do I find the nonzero eigenvectors of the complex eigenvalues?
 
Physics news on Phys.org
rayne said:
Matrix A:
0 -6 10
-2 12 -20
-1 6 -10

I got the eigenvalues of: 0, 1+i, and 1-i. I can find the eigenvector of the eigenvalue 0, but for the complex eigenvalues, I keep on getting the reduced row echelon form of:
1 0 0 | 0
0 1 0 | 0
0 0 1 | 0

So, how do I find the nonzero eigenvectors of the complex eigenvalues?

Hi rayne!

It means that 1+i and 1-i are not actually eigenvalues.
How did you conclude they were?
 
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

Replies
33
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
Replies
3
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
3K
Replies
1
Views
2K