I Orthogonality of Eigenvectors of Linear Operator and its Adjoint

[ughpleasenope]

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.

 

[Office_Shredder]

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.

 

[ughpleasenope]

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.

 

[Office_Shredder]

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)