1. Jan 7, 2008

mathboy

Let T:V -> V be a linear operator on a finite-dimensional inner product space V.
Prove that rank(T) = rank(T*).

So far I've proven that rank (T*T) = rank(T) by showing that ker(T*T) = ker(T). But I can't think of how to go from there.

2. Jan 8, 2008

mathboy

Oh, should I use the fact that the number of linearly independent rows of a matrix A is equal to the number of linearly independent columns of A*, and that rank(T) = rank of matrix represenation of T wrt to a basis? Or is there a better way?

rank(T)=rank([T])=rank([T]*)=rank([T*])=rank(T*)

Last edited: Jan 8, 2008
3. Jan 8, 2008

MathematicalPhysicist

well, dimV=dimT+dimKerT=dimT*+dimkerT*
but kerT=kerT*, one way to prove it is if
v not zero in ker T, then T(v)=0, so 0=<Tv,v>=<v,T*v>
but v isn't zero then T*(v)=0, you can do it also vice versa.

4. Jan 8, 2008

mathboy

But <v,T*v>=0 implies T*v=0 only if <v,T*v>=0 for ALL v in V, not just for v in kerT.

5. Jan 8, 2008

morphism

imT* is the orthogonal complement of kerT. So dimV=rankT*+dimkerT=rankT+dimkerT.

6. Jan 8, 2008

mathboy

Wow! Morphism, do you look these proofs up somewhere, or do you figure it out completely from scratch? If the latter, then you must be a genius!

7. Jan 8, 2008

morphism

It's just experience and not any form of genius - I already knew that fact, and it turned out to be helpful here.

8. Oct 4, 2010

Cassandra2182

How is the adjoint of the adjoint of a linear operator, the linear operator itself? i.e. A**=A?