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
vectors that span [math]\mathbb{R}^3[/math] are linearly independent, the rest is obvious
If you cannot solve this problem immediately you do not understand the definition of "span"
go reread it
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?
The 0 vectors don't span R^3.
What matrix with linearly independent columns would ever do that
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
Because you shouldve noticed that
[eqn]c_1 0 + c_2 0 + \ldots + c_n 0 = 0[/eqn]
>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
please send help i don't get what span or Rn is anymore when i thought i did
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
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.
So what? Read the question.
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.
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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.
This user 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
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.
203 at GMU?
caught you
I was only thinking of bases when I was posting this pls no bully
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
Every statistic is actually 50% because everything either happens or it doesn't.