Finding the eigenvectors of a 2nd multiplicity engenvalue

Click For Summary
SUMMARY

The discussion focuses on finding eigenvectors for the matrix [[2, 1, 0], [1, 2, 0], [0, 0, 3]]. The eigenvalues identified are 1 and 3, with 3 having a multiplicity of 2. The eigenvector corresponding to the eigenvalue 1 is [1, -1, 0], while for the eigenvalue 3, one eigenvector is [1, 1, 1]. The discussion highlights that any linear combination of eigenvectors satisfying the conditions x1=x2 and x3=x3 can also serve as valid eigenvectors, indicating the presence of a two-dimensional eigenspace for the eigenvalue 3.

PREREQUISITES
  • Understanding of eigenvalues and eigenvectors
  • Familiarity with matrix operations
  • Knowledge of linear combinations in vector spaces
  • Basic concepts of linear algebra
NEXT STEPS
  • Explore the concept of eigenspaces and their dimensions
  • Learn about linear combinations of vectors in linear algebra
  • Study the process of finding eigenvalues and eigenvectors using characteristic polynomials
  • Investigate the implications of multiplicity in eigenvalues on eigenvector solutions
USEFUL FOR

Students studying linear algebra, mathematicians interested in eigenvalue problems, and educators teaching matrix theory and its applications.

xdrgnh
Messages
415
Reaction score
0

Homework Statement


I'm given this matrice 2 1 0
1 2 0
0 0 3
and I need to find it's eigenvectors



Homework Equations





The Attempt at a Solution



So I get the eigenvalues to be 1,3,3 with 3 being the one with multiplicity of 2. For the eigenvector for 1 I get 1,-1,0 and for 3 I get 1,1,1 but here is the problem. For the other eigen vector for 3 the answer can be anything that satisfies x1=x2
x2=x1
x3=x3

so can something like 1,1,2 be the answer?
 
Physics news on Phys.org
xdrgnh said:

Homework Statement


I'm given this matrice 2 1 0
1 2 0
0 0 3
and I need to find it's eigenvectors



Homework Equations





The Attempt at a Solution



So I get the eigenvalues to be 1,3,3 with 3 being the one with multiplicity of 2. For the eigenvector for 1 I get 1,-1,0 and for 3 I get 1,1,1 but here is the problem. For the other eigen vector for 3 the answer can be anything that satisfies x1=x2
x2=x1
x3=x3

so can something like 1,1,2 be the answer?

Sure it could. Try it out if you are unsure. There are a lot of choices for specifying the eigenvectors. Any linear combination of 1,1,1 and 1,1,2 will also be an eigenvector with eigenvalue 3.
 
But wouldn't that mean there is no one definitive eigenbasis?
 
xdrgnh said:
But wouldn't that mean there is no one definitive eigenbasis?

Sure it would. There never is. You have a two dimensional space of eigenvectors with eigenvalue 3. There are lots of ways to choose a basis. I wouldn't call any of them 'definitive'.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K