Orthogonality of Eigenvectors of Linear Operator and its Adjoint

Click For Summary

Discussion Overview

The discussion revolves around the properties of eigenvectors of a linear operator and its adjoint in the context of a finite-dimensional complex vector space with a Hermitian inner product. Participants explore the conditions under which a linear operator is diagonalizable and the relationship between the eigenvectors of the operator and its adjoint.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant proposes proving that a linear operator T is diagonalizable if and only if for every eigenvector v of T, there exists an eigenvector u of T^* such that their inner product is non-zero.
  • Another participant agrees that the direction assuming T is diagonalizable is straightforward but expresses uncertainty about the reverse direction, suggesting it sounds plausible but requires further thought.
  • A later reply requests elaboration on the implications of T being diagonalizable and questions what kind of basis for the dual space V^* can be derived from this situation.

Areas of Agreement / Disagreement

Participants generally agree on the straightforwardness of one direction of the proof but express uncertainty and seek clarification regarding the other direction, indicating that the discussion remains unresolved.

Contextual Notes

Participants mention generalized eigenvectors and the nature of bases in relation to the operator and its adjoint, but these concepts remain underexplored and not fully resolved within the discussion.

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.
 
  • Like
Likes   Reactions: 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)
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 13 ·
Replies
13
Views
4K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K