**Definition. **A vector space is *finite dimensional* if it has a basis consisting of a finite number of vectors. The unique number of vectors in each basis for is called the *dimension* of and is denoted by .

**Definition. **Let and be vector spaces and let be linear. The *null space* (or kernel) of is the set of all vectors such that ; that is, . The *range* (or image) of is the subset of consisting of all images, under , of vectors in ; that is, .

**Definition. **Let and be vector spaces and let be linear. The *nullity *of is the dimension of denoted by . The* rank* of is the dimension of denoted by .

**Rank Nullity Theorem. **Let and be vector spaces and let be linear. If is finite dimensional, then .

*Proof.* Suppose that forms a basis for . We can extend this to form a basis for , . Since and , it suffices to show that . I claim that forms a basis for .

First we show that spans . Let . Then there exist unique scalars and such that

.

Thus,

.

Since each is in , we have that and so . This shows that spans .

Next we show that is linearly independent. Suppose that

for scalars . Since is linear, we have that

.

Thus, is in . Since span , there exist scalars such that

.

Which implies that

.

But, is a basis for which means that all the and must be zero. Thus, is linearly independent and is indeed a basis for . This proves that as desired.

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