Eigenvalues and eigenvectors of symmetric 2x2 matrix?

Click For Summary
A 2x2 symmetric matrix can be diagonalized, but it does not necessarily have an eigenvalue of 0. The trace of a symmetric matrix is equal to the sum of its eigenvalues, which may have led to some confusion. Eigenvalues of any symmetric matrix are guaranteed to be real numbers. A 2x2 symmetric matrix will have eigenvalues equal to its trace and 0 only if its determinant is zero. The discussion clarifies that not all symmetric matrices conform to the initial assumption regarding their eigenvalues.
malawi_glenn
Science Advisor
Messages
6,735
Reaction score
2,434
Hello

I recall, I think, that there is a lemma which states that a 2x2 symmetric matrix can be diagonalized so that its eigenvalues are (trace) and 0.

I can not find it anywhere =/ I think it was a physics teacher who told us this a couple of years ago, can anyone enlighten me?

cheers
 
Physics news on Phys.org
A general 2x2 symmetric matrix will not have an eigenvalue 0. The trace of a symmetric 2x2 matrix is equal to the sum of it's eigenvalues, maybe that was what you were thinkig of?
 
Last edited:
dx said:
A general 2x2 symmetric matrix will not have an eigenvalue 0. The trace of a symmetric 2x2 matrix is equal to the sum of it's eigenvalues, maybe that was what you were thinkig of?

hmm yeah, maybe something like that.

Thanx for input
 
The trace of a matrix is always equal to the sum of it's eigenvalues. I don't think there is anything special about the eigenvalues of a 2x2 symmetric matrix, but eigen values of any symmetric matrix will be real (non-imaginary).
 
The eigenvalues of any 2x2 matrix will be its trace and 0 if and only if the determinant of the matrix vanishes
 
<<I recall, I think, that there is a lemma which states that a 2x2 symmetric matrix can be diagonalized so that its eigenvalues are (trace) and 0>>

We can see that is not true by considering a trivial example of a matrix that already is diagonalized.

For example, if you digonalize an identity matrix, you get back the identity matrix. It, of course, does not have (trace) and zero as the diagonal elements.
 
I am studying the mathematical formalism behind non-commutative geometry approach to quantum gravity. I was reading about Hopf algebras and their Drinfeld twist with a specific example of the Moyal-Weyl twist defined as F=exp(-iλ/2θ^(μν)∂_μ⊗∂_ν) where λ is a constant parametar and θ antisymmetric constant tensor. {∂_μ} is the basis of the tangent vector space over the underlying spacetime Now, from my understanding the enveloping algebra which appears in the definition of the Hopf algebra...

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 33 ·
2
Replies
33
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
6K
  • · Replies 2 ·
Replies
2
Views
9K
  • · Replies 5 ·
Replies
5
Views
2K