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^{n}. If k < n then the set does

*not*span ℝ

^{n}. Why is this? Are there vectors in ℝ

^{n}that aren't combinations of the vectors in ##V##?

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Take numbers. If you want to span ##\mathbb{R}^3##, can it be done by two vectors: length and width? This only applies for the case ##k < n##. If we have ##n## or more vectors, we cannot say anything, because in this case, e.g. they could all be different multiples of just one vector, and then they still just span a line, although they might be many. But if we have less, then there is definitely something missing. In the example the height.^{n}. If k < n then the set doesnotspan ℝ^{n}. Why is this? Are there vectors in ℝ^{n}that aren't combinations of the vectors in ##V##?

- #3

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Take numbers. If you want to span ##\mathbb{R}^3##, can it be done by two vectors: length and width?

So, vectors in ℝ

If we have a set of two vectors, such as ##S = \left\{\begin{bmatrix} 1 \\ 0 \\ 0\end{bmatrix},\begin{bmatrix} 0 \\ 1 \\ 0\end{bmatrix}\right\}##, then the vector ##\begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix}## would

That right?

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All linear combinations of vectors in ##S## have the third component zero. They span the ##(x,y)## plane, but only the plane at height ##z=0##.

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Now, what happens if none of the components of each vector are zero? I mean, I'm sure they still don't span ℝ

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See, to answer the question, it first must be clear what has to be shown. A proof could be: Assume ##\mathbb{R}^3## is (linearly) spanned by two vectors ##v_1,v_2\in \mathbb{R}^3##. Then all ##w \in \mathbb{R}^3## can be written as ##w=\alpha v_1 + \beta v_2##. The goal is to show, that there is a vector in ##\mathbb{R}^3## which cannot be written this way. Therefore the question "What is ##\mathbb{R}^3##? has to be answered first. If we take the triplet definition, then a proof will result in a row echelon procedure applied on ##v_1,v_2,w## and the choice of the third component of ##w## which isn't reached by the choices of ##\alpha, \beta##.

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Ok, how about this, as anargument for why 2 vectors ## v_1,v_2 ## are not enough:

1) Find the cross froduct ##w:= v_1 \times v_2 ##. Then w is perpendicular to ##v_1, v_2##; both ##v_1, v_2## are in the plane normal to ##w## passing through

##v_1, v_2##. For any two points there is a unique plane going through the two points.

2) Consider the sum ##c_1v_1+c_2v_2 ##. This is a parallelogram (EDIT: Remember how the sum of vectors is defined, re the parallellogram rule/law), contained in a plane normal to ##v_1, v_2 ## . This sum will therefore not "escape" outside of the plane.

1) Find the cross froduct ##w:= v_1 \times v_2 ##. Then w is perpendicular to ##v_1, v_2##; both ##v_1, v_2## are in the plane normal to ##w## passing through

##v_1, v_2##. For any two points there is a unique plane going through the two points.

2) Consider the sum ##c_1v_1+c_2v_2 ##. This is a parallelogram (EDIT: Remember how the sum of vectors is defined, re the parallellogram rule/law), contained in a plane normal to ##v_1, v_2 ## . This sum will therefore not "escape" outside of the plane.

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I think one can use the general result (if one is allowed to do so) thatNow, what happens if none of the components of each vector are zero? I mean, I'm sure they still don't span ℝ^{3}, but I'm unsure as how to go about proving it.

A basis B for W means a finite set of vectors in W so that:

-- B is linearly independent

-- span(B)=W

Now if one takes the subspace W as equal to ℝ

Any two different bases for ℝ

========

But this leaves the possibility that there could be some another set which is

However, for any such vector A there will be another set B such that:

-- B⊂A

-- B is linearly independent

-- span(B)=span(A) (which is just equal to ℝ

This means that B is a basis for ℝ

=============

More directly (probably a somewhat less rigorous phrasing of post#7), one can use the reasoning for ℝ^3 that any two linearly independent vectors uniquely identify a plane (and if they aren't then they just uniquely identify a line going through origin in ℝ^3). So the set of all linear combinations will be either:

-- a line going through origin in ℝ^3

-- a plane (that also includes the origin) in ℝ^3

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Now, what happens if none of the components of each vector are zero? I mean, I'm sure they still don't span ℝ^{3}, but I'm unsure as how to go about proving it.

If you want a geometric argument, then any two vectors lie in a plane. All linear combinations of these two vectors must lie on the same plane, so cannot span all of ##\mathbb{R}^3##.

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Great minds think alike ; ) , see my #7.If you want a geometric argument, then any two vectors lie in a plane. All linear combinations of these two vectors must lie on the same plane, so cannot span all of ##\mathbb{R}^3##.

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1) Find the cross froduct ##w:= v_1 \times v_2 ##. Then w is perpendicular to ##v_1, v_2##; both ##v_1, v_2## are in the plane normal to ##w## passing through

##v_1, v_2##. For any two points there is a unique plane going through the two points.

2) Consider the sum ##c_1v_1+c_2v_2 ##. This is a parallelogram (EDIT: Remember how the sum of vectors is defined, re the parallellogram rule/law), contained in a plane normal to ##v_1, v_2 ## . This sum will therefore not "escape" outside of the plane.

Thanks WWGD. The geometric visualization makes sense. I can see that any two vectors in ℝ

While I can't take the visualization any further than ℝ

@SSequence thanks for the explanation. I can see now that any basis for ℝ

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Yes; take any point ##p## you know is in the plane, and translate it by any non-zero amount. The point will not be in the plane. So you can form an infinite collection ##\{p+ c \} ## where ##p ## is in the plane and ##c \in \mathbb R^3-\{0\} ## where ##0## is the 0 vector in ##\mathbb R^3## EDIT And notice you get an infinite collection _for each ##p##_ in the plane. Actually, in a sense, "most" points by many measures, are _not_ in the plane.Thanks WWGD. The geometric visualization makes sense. I can see that any two vectors in ℝ^{3}will have one vector that is perpendicular to them both, and thus ℝ^{3}will have a vector that's not a linear combination of the two vectors in the set. It seems to me that there are an infinite number of vectors outside of the plane formed by the span of ##S##, where ##S## is any two vectors in ℝ^{3}, right?

.

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Now, the standard basis in ##\mathbb R^n##: ##\{\mathbf e_1,\mathbf e_2,\dots \mathbf e_n\}##, is linearly independent, by definition, so every set of vectors in ##\mathbb R^n## which span ##\mathbb R^n## must have at least ##n## elements.

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This isn't true in general. If u = <1, 0, 1> and v = <2, 0, 2>, both vectors have an infinite number of vectors that are perpendicular to them. The two vectors I gave determine only a line, not a plane.Thanks WWGD. The geometric visualization makes sense. I can see that any two vectors in ℝ^{3}will have one vector that is perpendicular to them both

Drakkith said:, and thus ℝ^{3}will have a vector that's not a linear combination of the two vectors in the set. It seems to me that there are an infinite number of vectors outside of the plane formed by the span of ##S##, where ##S## is any two vectors in ℝ^{3}, right?

But, we could have a set of n + 1 vectors thatDrakkith said:Well, yes. The span of the vectors in your set S of post #3 determine a plane; namely, the x-y plane. Any vector that "pokes" up out of this plan (i.e., has a nonzero z-coordinate) won't lie in this plane.

While I can't take the visualization any further than ℝ^{3}, I can accept that this holds true for any ℝ^{n}.

@SSequence thanks for the explanation. I can see now that any basis for ℝ^{n}requires exactly n vectors. Any more and the set is not linearly independent. Any fewer an the set does not span ℝ^{n}.span##\mathbb R^n## (i.e., any vector in ##\mathbb R^n## is some linear combination of these n + 1 vectors, but as you point out, this set couldn't be a basis because there are too many of them -- the set must necessarily be linearly dependent. And we could have a set consisting of n - 1 vectors that is linearly independent, but there are too few to be a basis for ##\mathbb R^n##. However, this set spans a subspace of ##\mathbb R^n## of dimension n - 1

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But this is not part of the argument, the argument is that once you find the cross-product you do define a unique plane containing the two vectors and you can show that the sum (and linear combinations) stay in the plane. EDIT: But what you say is indeed correct, two points determine only a unique line and this line can be contained in infinitely-many planes. And, yes, you do need linear independence of the vectors.This isn't true in general. If u = <1, 0, 1> and v = <2, 0, 2>, both vectors have an infinite number of vectors that are perpendicular to them. The two vectors I gave determine only a line, not a plane.

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I understand that. I was responding to what Drakkith said, about "any two vectors in ##\mathbb R^n## ..."But this is not part of the argument, the argument is that once you find the cross-product you do define a unique plane containing the two vectors and you can show that the sum (and linear combinations) stay in the plane.

WWGD said:EDIT: But what you say is indeed correct, two points determine only a unique line and this line can be contained in infinitely-many planes. And, yes, you do need linear independence of the vectors.

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Sorry, I should have said two linearly independent vectors.

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This thread would have been simpler if it had started with the example of one vector in R

In the general case of R

PS. Unfortunately, it is not clear to me how to rigorously come up with a specific vector that is not in the space spanned by the k vectors. Maybe someone else can (or already has) given a method.

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This thread would have been simpler if it had started with the example of one vector in R^{2}. Clearly a single vector can only generate a line and not all of R^{2}.

In the general case of R^{n}and k vectors (k < n), the Gram-Schmidt orthogonalization process can turn the original set of vectors into a set of basis vectors that essentially spans R^{k}. So it is intuitive to see that the spanned space is not the same as R^{n}.

PS. Unfortunately, it is not clear to me how to rigorously come up with a specific vector that is not in the space spanned by the k vectors. Maybe someone else can (or already has) given a method.

Good point; Gram-Schmidt process preserves the span.

I think you can use the fundamental theorem of linear algebra : https://en.wikipedia.org/wiki/Fundamental_theorem_of_linear_algebra

EDIT to find element that is not in row space.

An element in the kernel of the operator will be in the perp space of the Row space. Since ## A^{\perp} \cap A=\{0\} ## * , just find a non-zero element in the kernel, since kernel of map is the complementary subspace. Note you can also use this to extend basis of subspace into basis for space.

* This is true in "most cases"; I think there are symplectic, other spaces where it is not true.

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PS. Unfortunately, it is not clear to me how to rigorously come up with a specific vector that is not in the space spanned by the k vectors. Maybe someone else can (or already has) given a method.

In some ways what follows is just a recut of what @WWGD said. However, the following two approaches are a bit more algorithmic and have a different sort of feel. Here are two simple and general approaches.

1.) using ##A\mathbf x = \mathbf b##, march through any valid basis -- I'd suggest the standard basis, and set ##\mathbf b := \mathbf e_j ##. You have to run through this 'for loop' at most ##n## times. During each iteration, run Gaussian elimination and you

2.) Here is a sketch of a useful and elegant approach: suppose you have ##n-1## vectors that are linearly independent in some ##n## dimensional vector space. Generate an ##n## vector at random (uniform over ##[-1,1]## if in reals) and with probability one you have generated a linearly independent vector. You may confirm this by appending the random vector to the others and computing a determinant or running Gaussian elimination.

Note: this approach also has appeal over finite fields. If, say, you are in ##GF(2)## your random vector over ##\{0,1\}## will be linearly independent with at worst probability ##0.5##. Run ##r## independent trials and the probability that at least one of the random vectors is linearly independent is ##1 - (0.5)^r##. This approach actually works quite well, btw, when dealing with very large n. E.g. if ##n = 1,000## you could simply run the experiment with ##r = 20## and get at least one vector that is linearly independent with absurdly high probability. Of course, if you only have ##n-2## vectors that are linearly independent, the probability that your random vector is linearly independent only gets better. (And ##n-3## is even better from there, and so on.)

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A small caveat to this argument: the translation must be done in the "right way" , so that it is not done by adding ##c## where ##c## is a vector in the plane.Yes; take any point ##p## you know is in the plane, and translate it by any non-zero amount. The point will not be in the plane. So you can form an infinite collection ##\{p+ c \} ## where ##p ## is in the plane and ##c \in \mathbb R^3-\{0\} ## where ##0## is the 0 vector in ##\mathbb R^3## EDIT And notice you get an infinite collection _for each ##p##_ in the plane. Actually, in a sense, "most" points by many measures, are _not_ in the plane.

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an abstract proof that a vector space containing n independent vectors cannot be spanned by fewer than n vectors is given in the notes below, on page 16 and following, using the famous exchange argument of Riemann, usually attributed to Steinitz.

http://alpha.math.uga.edu/%7Eroy/laprimexp.pdf

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