Linear Transformation and Null Space

[Random Variable]

Homework Statement

Let (u,v,w) be a basis for vector space V, and let L be a linear transformation from V to vector space W. If (L(u),L(v),L(w)) is linearly dependent, then dim(Null Space(L)) > 1.

Homework Equations

The Attempt at a Solution

I don't see why dim(Null Space(L)) must be strictly greater than 1.


My proof that the dim(Null Space(L))>=1:

Let x be a vector in V that is in the Null Space of L.

Then L(x) = L(αu+βv+γw) = 0

L(αu)+L(βv)+L(γw) = 0

αL(u)+βL(v)+γL(w) = 0

Since the family (L(u),L(v),L(w)) is linearly dependent, α, β,, and γ are not all zero.

Therefore, Null Space(L) ≠ {0}, which implies that the dim(Null Space(L)) is at least 1.

 

[yyat]

You are right, it should be dim(Null Space(L)) >= 1. (Try constructing an example where dim(Null Space(L))=1.)

A small remark concerning your proof: when you say "α, β,, and γ are not all zero" you should instead write that "α, β,, and γ can be chosen so that not all are zero", because they can be all 0 (for x the zero vector). Then you can infer that x can be nonzero by reading the last three lines backwards.

 

[Random Variable]

Revised Proof:

Let x be a vector in V that is in the Null Space of L.

Then L(x) = L(αu+βv+γw) = 0

L(αu)+L(βv)+L(γw) = 0

αL(u)+βL(v)+γL(w) = 0

Since the family (L(u),L(v),L(w)) is linearly dependent, α,β,and γ may not all be zero.

Therefore, Null Space(L) consists of more than just the zero vector, which implies dim(Null Space(L)) is not zero.