Undergrad Orthogonality of Eigenvectors of Linear Operator and its Adjoint

Click For Summary
In a finite-dimensional complex vector space with a Hermitian inner product, the discussion focuses on proving that a linear operator T is diagonalizable if and only if each eigenvector v of T has a corresponding eigenvector u of its adjoint T* such that their inner product <u, v> is non-zero. The straightforward direction assumes T is diagonalizable, allowing for the construction of a basis of V consisting of eigenvectors of T. The challenge lies in demonstrating the converse, which is considered plausible but not immediately clear. Participants express a need for clarification on the implications for the dual space V* and its basis in relation to the diagonalization of T. The conversation highlights the complexities involved in understanding the relationship between T and T* in the context of eigenvectors.
ughpleasenope
Messages
2
Reaction score
0
Suppose we have V, a finite-dimensional complex vector space with a Hermitian inner product. Let T: V to V be an arbitrary linear operator, and T^* be its adjoint.

I wish to prove that T is diagonalizable iff for every eigenvector v of T, there is an eigenvector u of T^* such that <u, v> is not equal to 0.

I've been thinking about generalized eigenvectors, but have not really gotten anywhere.
 
Physics news on Phys.org
The direction where you assume T is diagonalizable is pretty straightforward I think?

The other direction is not immediately obviously true to me but sounds plausible, I'll sleep on it.
 
Reply
  • Like
Likes ughpleasenope
Office_Shredder said:
The direction where you assume T is diagonalizable is pretty straightforward I think?

The other direction is not immediately obviously true to me but sounds plausible, I'll sleep on it.
Would you mind elaborating? I've struggled with this for a while.
 
If T is diagonalizable, then you can write down a basis of V which are all eigenvectors of T.

What kind of basis of ##V^*## do you get from this? (I guess if your class is very matrix based this question might not make sense)
 

Thread 'Confusion about the Moyal-Weyl twist'

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...
View full post »

Similar threads

Undergrad Does Axler's Spectral Theorem Imply Normal Matrices?
  • · Replies 3 ·
Replies
3
Views
2K
Undergrad The Orthogonality of the Eigenvectors of a 2x2 Hermitian Matrix
  • · Replies 13 ·
Replies
13
Views
3K
Undergrad Proof that if T is Hermitian, eigenvectors form an orthonormal basis
  • · Replies 3 ·
Replies
3
Views
3K
Undergrad If T is diagonalizable then is restriction operator diagonalizable?
  • · Replies 3 ·
Replies
3
Views
2K
Undergrad Question about an eigenvalue problem: range space
  • · Replies 1 ·
Replies
1
Views
2K
Graduate Eigenvectors and matrix inner product
  • · Replies 1 ·
Replies
1
Views
3K
Undergrad If L is diagonalizable, algebraic & geometric multiplicities are equal
  • · Replies 4 ·
Replies
4
Views
3K
Undergrad Question from a proof in Axler 2nd Ed, 'Linear Algebra Done Right'
  • · Replies 3 ·
Replies
3
Views
3K
Undergrad Proof by induction of block diagonal decomposition of a matrix
  • · Replies 4 ·
Replies
4
Views
2K
High School Is the Null Space of an Operator Defined by Its Zero Eigenvalue?
  • · Replies 4 ·
Replies
4
Views
5K