another proof of rank-nullity theorem

Let ϕ:VW be a linear transformation from vector spacesMathworldPlanetmath V to W. Recall that the rank of ϕ is the dimensionMathworldPlanetmathPlanetmath of the image of ϕ and the nullityMathworldPlanetmath of ϕ is the dimension of the kernel of ϕ.

Proposition 1.



Let K=ker(ϕ). K is a subspaceMathworldPlanetmathPlanetmath of V so it has a unique algebraic complement L such that V=KL. It is evident that


since K and L have disjoint bases and the union of their bases is a basis for V.

Define ϕ:Lϕ(V) by restrictionPlanetmathPlanetmath of ϕ to the subspace L. ϕ is obviously a linear transformation. If ϕ(v)=0, then ϕ(v)=ϕ(v)=0 so that vK. Since vL as well, we have vKL={0}, or v=0. This means that ϕ is one-to-one. Next, pick any wϕ(V). So there is some vV with ϕ(v)=w. Write v=x+y with xK and yL. So ϕ(y)=ϕ(y)=0+ϕ(y)=ϕ(x)+ϕ(y)=ϕ(v)=w, and therefore ϕ is onto. This means that L is isomorphicPlanetmathPlanetmathPlanetmath to ϕ(V), which is equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath to saying that dim(L)=dim(ϕ(V))=rank(ϕ). Finally, we have


Remark. The dimension of V is not assumed to be finite in this proof. For another approach (where finite dimensionality of V is assumed), please see this entry (

Title another proof of rank-nullity theorem
Canonical name AnotherProofOfRanknullityTheorem
Date of creation 2013-03-22 18:06:14
Last modified on 2013-03-22 18:06:14
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 4
Author CWoo (3771)
Entry type Proof
Classification msc 15A03
Related topic ProofOfRankNullityTheorem