Adjoint of a linear operator (2)

[kingwinner]

Q: Suppose V is a finite dimensional inner product space and T:V->V a linear operator.
a) Prove im(T*)=(ker T)^(|)
b) Prove rank(T)=rank(T*)

Note: ^(|) is orthogonal complement

For this question, I don't even know how to start, so it would be nice if someone can give me some hints. Thank you!

 

[HallsofIvy]

T* is the linear transformation such that, for all u and v, <Tu,v>= <u,T*v> . ker T= {v, Tv= 0). The "orthogonal complement" of ker T is {u |<u, v>= 0 for all v in ker T}. If w is in the image of T* then w= T*v for some v. Then <u, w>= <u,T*v>= <Tu, v> and if u is in the kernal of T, that is equal to what? That proves that im(T*) is a subset of the orthogonal complement of ker T. Prove the other way similarly. Of course, you should know that the rank(T)+ nullity(T)= dimension of V. (Nullity(T)= dimension of kernal of T and rank(T)= dimension of im(T).