Rank Nullity Theorem
The Rank Nullity Theorem is a result in linear algebra. To provide some background to understand the full claim, let and be vector spaces and let be a linear transformation. We define where . Additionally, we define where . In other words, rank is the dimension of the linear transformation's image, and nullity is the dimension of it's kernel (the word nullity comes from the term "null space" which is used to describe the physical subset of the domain of the linear transformation that maps to the zero vector). The theorem states the following:
Proof: Let and be vector spaces and let be linear. We first claim that if is a basis of , then
By definition we have . Because is a subspace (this is easy to verify), then the basis of is contained in or that it contains
Now we must prove that contains in order for . To do this, let . Then for some . Because is a basis for , we can use the linearity of to get
so is contained in its basis, which means that the two must be equal, done. With this corollary out of the way, we head to the main course. Obviously this is under the assumption that is finite, so assume that and such that is a basis for . We can then extend this basis into for . We seek to show that is a basis for , because that would mean which would prove what we want.
To show that is a basis, we have two things to do. We need to show that generates and that is linearly independent, which would satisfy both criteria for a basis. The first part is easy, because since we have for each we get
Now we have to show that is linearly independent. Using the linearity of , we would have
so we can conclude that . Then by taking a set of scalars we get
Since is a basis of , we see that this implies that is linearly independent, completing our work.