linear algebra

[Anonymous No.8389808]

anyone know linear algebra? i have to prove that

If columns of A span R^3, then AX = B has a solution for every B.

where A = [U V W] and X = [x y z]. and U, V and W are elements in R^3

i'm lost. reading the textbook for the class which is linear algebra and its applications by david lay and I'm reading about vectors and spans and stuff but i don't see how to translate what i'm reading into solving this

[Anonymous No.8389814]

vectors that span [math]\mathbb{R}^3[/math] are linearly independent, the rest is obvious

[Anonymous No.8389816]

>>8389808
If you cannot solve this problem immediately you do not understand the definition of "span"
go reread it

[Anonymous No.8389820]

>>8389814
but couldn't it be
[0 0 0 b1]
[0 0 0 b2]
[0 0 0 b3]
and then there's no solution when one of the b's isn't equal to 0 because it'd be inconsistent?

[Anonymous No.8389827]

>>8389820
The 0 vectors don't span R^3.

[Anonymous No.8389830]

>>8389820
What matrix with linearly independent columns would ever do that

[Anonymous No.8389836]

>>8389827
my book doesn't say that, it just says the vector with all zero entires is called the zero vector. it doesn't say they're not in R

[Anonymous No.8389841]

>>8389836
Because you shouldve noticed that
[eqn]c_1 0 + c_2 0 + \ldots + c_n 0 = 0[/eqn]

[Anonymous No.8389847]

>>8389836
>it doesn't say they're not in R
It is in R^3.

It doesn't span R^3, which is a different thing.

A vector or a set of vectors spanning a set means that you can write a vector of the spanned set as a linear combination of those vectors/vector

[Anonymous No.8389872]

please send help i don't get what span or Rn is anymore when i thought i did

[Anonymous No.8389878]

>>8389872
go read the fucking definition then
why should /sci/ spoonfeed you stuff that's literally copy-pasted from the formal definition in your book

[Anonymous No.8389882]

>>8389878
my good friends on /g/ were right, you guys aren't helpful at all

[Anonymous No.8389886]

>>8389872
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]

>>8389882
Nobody can help you if you aren't willing to help yourself.

[Anonymous No.8389889]

>>8389886
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?

[Anonymous No.8389897]

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.

[Anonymous No.8389899]

>>8389889
>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

[Anonymous No.8389906]

>>8389889
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.

[Anonymous No.8389933]

>>8389906
>>8389899
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?

[Anonymous No.8389956]

>>8389933
>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.

[Anonymous No.8389965]

>>8389956
thank you, i think i have it now

[Anonymous No.8390063]

What's this thing that the spanning vectors must be independent? What's required is just that any vector in the vector space can be expressed as a linear combination of these. To require linear independence in addition is to require they are a basis for the vector space.

[Anonymous No.8390069]

>>8390063
So what? Read the question.

[Anonymous No.8390077]

>>8390069
>>8389814
But this is false. We have [math]\mathbb{R}^3 = \langle \mathbb{R}^3 \rangle[math] as a counter example. Since OP is clearly pretty lost, I don't think he should be misinformed like this.

[Anonymous No.8390080]

>>8390069 #
>>8389814 #
But this is false. We have [math]\mathbb{R}^3 = \langle \mathbb{R}^3 \rangle[/math] as a counter example. Since OP is clearly pretty lost, I don't think he should be misinformed like this.

[Anonymous No.8390089]

>>8390080

This anon >>8389814 may be a retard but the question mentions only three vectors that also span R^3, because its dimension is 3, it does follow that the vectors are linearly independent.

Also kys weeb faggot

[Anonymous No.8390119]

>>8390089
That's probably what he means. Yet, he didn't say "any n vectors spanning [math]\mathbb{R^n}[/math]", but "any vectors", and the latter is false. That's the problem in his post.

[Anonymous No.8390146]

>>8389808
203 at GMU?

caught you

[Anonymous No.8390165]

>>8390119
I was only thinking of bases when I was posting this pls no bully

[Anonymous No.8390247]

If A is a 3x3 matrix, and the columns of A span R3, then they are linearly independent.
This means that the determinant of the matrix is nonzero.
This means you can invert the matrix.
This means you can solve for X: X = inverse(A)*B

[Anonymous No.8390263]

Every statistic is actually 50% because everything either happens or it doesn't.