another proof of rank-nullity theorem
since and have disjoint bases and the union of their bases is a basis for .
Define by restriction of to the subspace . is obviously a linear transformation. If , then so that . Since as well, we have , or . This means that is one-to-one. Next, pick any . So there is some with . Write with and . So , and therefore is onto. This means that is isomorphic to , which is equivalent to saying that . Finally, we have
Remark. The dimension of is not assumed to be finite in this proof. For another approach (where finite dimensionality of is assumed), please see this entry (http://planetmath.org/ProofOfRankNullityTheorem).
|Title||another proof of rank-nullity theorem|
|Date of creation||2013-03-22 18:06:14|
|Last modified on||2013-03-22 18:06:14|
|Last modified by||CWoo (3771)|