You

**said** "Also assume {

} forms a basis for ker L.

We can extend this to form a basis for V: {

}".

Do you not understand what "extend" means? We have a basis for ker L, a subspace of V and we add

**new** vectors from V, that are

**not in** ker L to fill out a full basis for V.

You also say, in your first post, "I think I see why {

} is a basis for ker L'. No, that's impossible (unless ker L= V, of course) because we have

**extended** adding vectors

, etc. vectors that are NOT in ker L.

Here's how you do that: if ker L= V, then we are done. A basis for one is a basis for the other.

If ker L is not V, then there exist vectors in V that are not in ker L. Pick any one and add it to the basis of ker L. If we call that new vector

, then we now have {

}. Since

was not in

L, it cannot be written as a linear combination of the u vectors so this set is still a set of independent vectors. If it spans V, we are done. If it does not, by definition of "span", there must exist a vector,

in V that is not in the span. Add it to the set. Again, because it was not in the span of the previous vectors, it is independent of them. See it this new set spans V. If it does, we are done- we have a set that is both independent and spans V- a basis for V. If not, there exist some other vector,

that is not in its span. Add that vector.

That process must eventually end because we are given that V is a finite dimensional vector space. There must be some number N such that no set of more than N vectors can be independent.