kerL={0} if and only if L is injective


A linear map between vector spacesMathworldPlanetmath is injective if and only if its kernel is {0}.


Let L:VW be a linear map. Suppose L is injective, and L(v)=0 for some vector vV. Also, L(0)=0 because L is linear. Then L(v)=L(0), so v=0. On the other hand, suppose kerL={0}, and L(v)=L(v) for vectors v,vV. Hence L(v-v)=L(v)-L(v)=0 because L is linear. Therefore, v-v is in kerL={0}, which means that v-v must be 0. ∎

Title kerL={0} if and only if L is injective
Canonical name operatornamekerL0IfAndOnlyIfLIsInjective
Date of creation 2013-03-22 14:44:46
Last modified on 2013-03-22 14:44:46
Owner Mathprof (13753)
Last modified by Mathprof (13753)
Numerical id 11
Author Mathprof (13753)
Entry type Theorem
Classification msc 15A04