Undergrad Orthogonality of Eigenvectors of Linear Operator and its Adjoint

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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
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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.
 
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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.
 
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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)
 
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