Solving Linear Systems with Hermitian Matrices

Click For Summary
SUMMARY

This discussion focuses on solving linear systems derived from Hermitian matrices, specifically addressing the eigenvalue problem with a given eigenvalue of -3. The user presents two equations, 5x+(3-i)y=0 and (3+i)x+2y=0, which arise from substituting the eigenvalue into the matrix. The user initially struggles with solving the system but realizes that setting ax = by allows for a straightforward solution approach. This method is crucial for tackling more complex equations in linear algebra.

PREREQUISITES
  • Understanding of Hermitian matrices and their properties
  • Familiarity with eigenvalues and eigenvectors
  • Basic knowledge of complex numbers and their manipulation
  • Experience with solving linear equations
NEXT STEPS
  • Study the properties of Hermitian matrices in depth
  • Learn techniques for finding eigenvalues and eigenvectors of matrices
  • Explore methods for solving linear systems involving complex coefficients
  • Investigate advanced linear algebra topics, such as the Singular Value Decomposition (SVD)
USEFUL FOR

Students and professionals in mathematics, particularly those studying linear algebra, as well as anyone working with Hermitian matrices and complex linear systems.

tamtam402
Messages
199
Reaction score
0

Homework Statement


I can find my eigenvalues just fine, and they're both real, as expected. My first eigenvalue is -3, which I know is correct.

I have the equations 5x+(3-i)y=0, (3+i)x+2y=0

Both of the equations come from my hermitian matrix, after I substituted λ=-3.

Homework Equations





The Attempt at a Solution



I have absolutely no idea how to solve this. This case is simple enough to be solved by trial and error, but how would I proceed if I had harder equations?

I can't use both equations since I get x=x or y=y if I substitute one into the other, since they're both the same equation.
 
Physics news on Phys.org


Nevermind, doing another (easier) exercise allowed me to see that I only have to set ax = by, and force a value to either x or y.
 

Similar threads

Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 31 ·
2
Replies
31
Views
4K
  • · Replies 13 ·
Replies
13
Views
11K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
2
Views
1K