Linear algebra

go read the fucking definition then
why should Veeky Forums spoonfeed you stuff that's literally copy-pasted from the formal definition in your book

my good friends on /g/ were right, you guys aren't helpful at all

Do a set of vectors [math]\{ v_1, v_2, v_3 \}[/math] span [math]\mathbb{R}^3[/math]?
In other words, are there weights [math]c_1, \, c_2, \, c_3[/math] such that
[eqn]c_1 v_1 + c_2 v_2 + c_3 v_3 = \begin{bmatrix} x \\ y \\ z\end{bmatrix}[/eqn]

If a some "formula" of [math]c_1, \, c_2, \, c_3[/math] exist and work, then your set of vectors are said to span [math]\mathbb{R}^3[/math]

Nobody can help you if you aren't willing to help yourself.

If the columns of A span R3, then no vector in the set can be defined as a linear combination of another, and they aren't zero vectors. Therefore, there can't be a pivot in the last column of augmented form of AX = B, so AX = B has a solution for every B.
>am i getting closer?

ayy yo hol up. if there is a stupid question general, then this should be posted there, if not, then the OP should have created one.

sorry OP I didn't mean to call you stupid.

>If the columns of A span R3, then no vector in the set can be defined as a linear combination of another, and they aren't zero vectors
Correct. Are you able to prove why a set of vectors containing the zero vector is linearly dependent though? (This may be relevant for your course)

>Therefore, there can't be a pivot in the last column of augmented form of AX = B, so AX = B has a solution for every B.
>am i getting closer?
You're there, congrats

You're closer in some sense, in that you actually have a correct definition now.
However the problem is that you just pulled this
>Therefore, there can't be a pivot in the last column of augmented form of AX = B
straight out of your ass, because you know the endpoint and the starting point of the proof but not the reasoning, so you left the reasoning out and just glued them straight together.

Do you understand what the matrix equation AX = B means?
It means that xU + yV + zW = B.

Doesn't this look like what the other guy gave you for the definition of span?
If you can map that form of the matrix equation to the span equation, you can provide an explanation why the columns spanning R3 means the equation is soluble.

it would be that if there was a row [0 0 0 b], then the columns wouldn't span R3. because they would only form a 2 dimensional plane instead of the entire 3d space as they would if they spanned r3. is that right? i watched a video no youtube that visualized things. if its right how do i turn it into being acceptable in a proof?

>it would be that if there was a row [0 0 0 b], then the columns wouldn't span R3.
You've basically got a proof. You can make it a bit more rigorous by showing specifically and concretely why that row forces the vectors to not span R3 (either make up a vector you can't get or show that the columns are linear combinations of each other because of that row).

A somewhat simpler way of showing it, without resorting to the rank test, is simply saying that since {U, V, W} spans R3, by definition every vector B in r3 is a linear combination of those vectors.
So there are constants a b c such that
aU + bV + cW = B
and that's your solution vector.

It's very important to remember the exact formal definition when trying to prove things.

thank you, i think i have it now