Question

Linear algebra: Linear mappings (transformations):

I'm gonna use x' and 0' to denote vectors, and L is a linear mapping (linear transformation). Why does

\[0L(x \prime) = 0 \prime\]

(scalar 0 * L(x') = zero vector)

Answer 1

If it wasn't that way, it wouldn't be a linear map. It's one of the basic axioms of a vector space.

Answer 2

Do you have a more specific answer? I don't now why I'm struggling with this idea since it seems sort of "obvious" but I don't know which theorem or axiom is applied.

The closest thing for vector spaces (V) relating to 0 vectors is:
a) There exists a linear mapping O: R^n -> R^m such that L + O = L. In particular, O is the linear mapping defined as O(x') = 0' for all x' in R^n

b) For each linear mapping L, there exists a linear mapping (-L) with the property that L + (-L) = O. In particular, (-L)(x') is the linear mapping defined by (-L)(x') = -L(x')

Answer 3

The way I see it, is that since L is a linear map between two vector spaces, it must preserve multiplication and addition. However, in any vector space, multiplying a vector by 0 must give the zero vector since you're multiplying the individual vector components by 0 (in more detail, this is because vector spaces are fields with a few extra things). Since you're multiplying the components by 0, the components become 0, so you get the zero vector.

Does this help?

Answer 4

Oh ok I see what you mean. Yes that helps thanks :)

Answer 5

You're welcome.