proof of rank-nullity theorem
The images of a basis of will span , and hence is finite-dimensional. Choose then a basis of and choose preimages such that
Choose a basis of . The result will follow once we show that is a basis of .
Let be given. Since , by definition, we can choose scalars such that
Linearity of now implies that and hence we can choose scalars such that
Therefore span .
Next, let be scalars such that
By applying to both sides of this equation it follows that
and since are linearly independent that
as well, and since are also assumed to be linearly independent we conclude that
also. Therefore are linearly independent, and are therefore a basis. Q.E.D.
|Title||proof of rank-nullity theorem|
|Date of creation||2013-03-22 12:25:13|
|Last modified on||2013-03-22 12:25:13|
|Last modified by||rmilson (146)|