# An equation related to the dimension of linear operator

1. Dec 1, 2008

### boombaby

1. The problem statement, all variables and given/known data
Let V be a finite vector space, and A, B be any two linear operator. Prove that,
rank A = rank B + dim(Im A $$\cap$$ Ker B)

3. The attempt at a solution
Since rank A = dim Im A
dim(Im B)+ dim(Ker B)=dim V
dim(Im A + Ker B)=dim(Im A)+dim(Ker B)-dim(Im A $$\cap$$ Ker B)
It seems equivalent to prove that dim(Im A + Ker B)=dim V

it is obvious that Im A + Ker B $$\subseteq$$ V. So it should be true that V $$\subseteq$$ Im A + Ker B.

If I let Ker B=<e_1,...,e_k> and extend it to a basis of V=<e_1,..e_k, e_k+1,...e_n>. Choose any $$x=\sum^{n}_{i=1} x_{i}e_{i}=\sum^{k}_{i=1} x_{i}e_{i} + \sum^{n}_{i=k+1} x_{i}e_{i}$$. The first term is in Ker B, the second is not in Ker B and therefore should be in Im A. This is what confuses me a lot since it is not necessary to be true IMO...

What's your idea? Where I made the mistakes? Can you give me any hint to prove it? Thanks very much

2. Dec 2, 2008

### morphism

Are A and B arbitrary operators on V? If so this isn't true unless V={0}. For a counterexample, take A=0 and let B be any nonzero operator.

3. Dec 2, 2008

THanks!