Homework Help: Prove the inverse image is a subspace.

1. Feb 9, 2012

trap101

Let T: V-->W be a surjective linear transformation and let X be a subspace of W. Assume tat Ker(T) and X are finite dimensonal.

Prove that T^-1 (X) = {v | T(v) is in X} is a subspace of V.

Ok I absolutely suck at showing things are a subspace of something....I don't even know where to start.

Thanks for the help

2. Feb 9, 2012

Deveno

a "subspace" is a subset U of a vector space V that is a vector space in its own right, when the vector addtion of V is restricted to U and the scalar multiplication is restricted to U.

since the operation on U is really "the same one" as on V, there are certain axioms of the entire canon of axioms of a vector space we don't actually have to prove for U, because they hold just as they did for V.

for example, if u+v = v+u for all u,v in V, then it is certainly STILL true if we insist that u,v are in U (since U is a subset of V, so everything in U is also in V).

we have no guarantee that when we restrict vector addition to U, that for u,v in U, that u+v will still BE in U. that is, that "+" restricted to U is a legitimate binary operation on U.

we also have no guarantee that for a scalar c, that cu will still be in U (which has to be true for U to be a vector space).

finally, the empty set is NEVER a vector space (every vector space contains, at minimum, a 0-vector. that's as small as a vector space can get). so it behooves us to ensure that something is actually IN our "subspace" U.

so typically, to establish a subSET is a subSPACE, you need to show 3 things:

a)if u,v are in U, so is u+v (closure of vector addition)
b)if c is in F, and u is in U, then cu is in U (closure of scalar multiplication)
c)the 0-vector of V is in U (alternatively, U is non-empty).

3. Feb 9, 2012

jgens

How about showing closure under addition and scalar multiplication? The other vector space properties are inherited from the original space.