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## 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.