# 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).

but there are some things we "don't get for free".

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.