Understanding Eigenvectors: Troubleshooting and Verification

Click For Summary
SUMMARY

The discussion focuses on verifying eigenvectors for the matrix A=[[3,4],[3,2]] with eigenvalues -1 and 6. The user initially computes the eigenvectors as [1,-1] and [-4/3,1], but corrects the second eigenvector to [4/3,1] based on the equation 3x-4y=0. It is established that eigenvectors can be represented in multiple forms due to their scalar multiplicative nature, confirming that (4/3,1) and (4,3) represent the same eigenvector.

PREREQUISITES
  • Understanding of eigenvalues and eigenvectors
  • Familiarity with null space calculations
  • Knowledge of matrix operations and transformations
  • Basic linear algebra concepts
NEXT STEPS
  • Study the properties of eigenvectors and eigenvalues in linear algebra
  • Learn about null space and its significance in eigenvalue problems
  • Explore the concept of scalar multiplication in vector spaces
  • Investigate the application of eigenvectors in real-world scenarios, such as PCA (Principal Component Analysis)
USEFUL FOR

Students and professionals in mathematics, physics, and engineering, particularly those studying linear algebra and its applications in various fields.

awl2k
Messages
1
Reaction score
0
Hello people!

I am having a bit trouble with verifying my result when i compute the eigenvectors for the following matrix:

A=[[3,4],[3,2]]

I know for sure that the eigenvalues is respectively -1 and 6, so i start finding a solution for the following null spaces:

1) N(A--1I)=[[4,4|0],[3,3|0]]~[[1,1|0],[0,0|0]] => x1 = x2 so the the vector x2[1,-1] will be a solution and therefor the first eigenvector is [1,-1]

2) N(A-6I)=[[-3,4|0],[3,-4|0]]~[[1,-4/3|0],[0,0|0]] => x1=4/3 so this indicate that x2[-4/3,1] will be a solution to the null space, and therefor the second eigen vector i [-4/3,1].

However the result should be [1,-1] [4,3] respectively. What am i doing wrong?
 
Physics news on Phys.org
[-4/3,1] is not an eigenvector of 6. The correct eigenvector is [4/3,1] because the equation you get is 3x-4y=0.
 
Eigenvectors are always undermined up to a multiplicative factor, since if A*x = λ*x, then
A*(nx) = λ*(nx). So (4/3,1) and (4,3) are the same eigenvector.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
Replies
2
Views
1K
  • · Replies 12 ·
Replies
12
Views
3K
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K