Proof of Rank-Nullity Theorem

- is one-one is the zero subspace of is a basis of
- If is finite dimensional vector space then The equality holds if and only if

The other parts can be similarly proved. Part follows from the previous two parts. height6pt width 6pt depth 0pt

The proof of the next theorem is immediate from the fact that and the definition of linear independence/dependence.

which is equivalent to showing that is the span of

We now prove that the set is a linearly independent set. Suppose the set is linearly dependent. Then, there exists scalars, not all zero such that

Or which in turn implies So, there exists scalars such that

That is,

Thus for as is a basis of In other words, we have shown that the set is a basis of Now, the required result follows. height6pt width 6pt depth 0pt

we now state another important implication of the Rank-nullity theorem.

Suppose is onto. Then Hence, But then by the rank-nullity Theorem 14.5.3, That is, is one-one.

Now we can assume that is one-one and onto. Hence, for every vector in the range, there is a unique vectors in the domain such that Therefore, for every in the range, we define

That is, has an inverse.

Let us now assume that has an inverse. Then it is clear that is one-one and onto. height6pt width 6pt depth 0pt

A K Lal 2007-09-12