One of the important results in linear algebra is the rank nullity theorem. Here I am going to present a proof of it which is slightly less known. The reason I like this proof is because it ties together many concepts and results quite nicely, and also because I independently thought of it.

The theorem (as is well known) says that if are vector spaces with and a linear map then .

In this proof I will further assume that is finite dimensional with dimension . A more general proof can be found on wikipedia.

We start by fixing two bases of and and obtain a matrix of relative to these bases. (Each is a row matrix). Then our theorem basically translates to . We let and claim that .

Clearly if then and so so that each is orthogonal to . Hence . Conversely if then so that , i.e. following which .

Now it only remains to invoke the result for any subspace of an inner product space to conclude that . In other words .

you reduce the original problem to “our theorem basically translates to \text{rank} (A)+\text{nullity}(A)=n”. Is this step needs a proof?

@Shiv: Recall that rank is the dim of the column space and nullity is the dim of the solution space.