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Basis/Dimension -- Help please

  1. Nov 14, 2015 #1
    • moved to homework forum so homework template headers are missing
    Does this set span R_2? Is this set a basis for R_2? How many dimensions does it have?

    1 -4
    2 3
    -4 6

    This can be row reduced to
    1 0
    0 1
    0 0

    In row-reduced echelon form

    I couldn't find the formatting to format this into a matrix, but these are column vectors.
     
  2. jcsd
  3. Nov 14, 2015 #2

    Krylov

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    What do you think? Recall your definitions (of "span", "basis" and "dimension") and it should not be hard to answer this question. (BTW: Strictly speaking, the set of these two column vectors does not have any dimension, because the notion of "dimension" in an LA context applies only to linear (sub)spaces.)
     
  4. Nov 14, 2015 #3

    HallsofIvy

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    I, personally, dislike setting vectors into matrices. I would prefer to use the basic definitions. Also, it can be confusing whether you intend the vectors to be the rows or columns. Actually, I think it is more common to do it as columns but, here, since you specify [itex]R^2[/itex], it is clear you intend the rows, (1, -4), (2, 3), and (-4, 6). One of the things you should know is that if the vector space has dimension "n", then NO set containing more than n vectors can be independent. (And no set containing fewer than n vectors can span the space.)

    Here, you have three vectors while [itex]R^2[/itex] has dimension 2 so they cannot be independent and so cannot be a basis. The can still span the space. A set of vectors "spans" a space if and only if every vector in that space can be written as a linear combination of vectors in the spatce. Here, any vector in [itex]R^2[/itex] can be written as "(x, y)". Do there exist numbers, A, B, C, such that A(1, -4)+ B(2, 3)+ C(-4, 6)= (x, y). Can we find A, B, C such that A+ 2B- 4C= x and -4A+ 3B+ 6C= y for any x and y? If we multiply the first equation by 2 and add to the second (equivalent to your "row reduction") we cancel "A" and get 7B- 2C= 2x+ y. We now have only that single equation but that means that, given any value for B, we have C= (7/2)B- x- y/2. B can be any number, there are, in fact, an infinite number of such linear combinations (more than one is a consequence of this set NOT being independent). Yes, this set spans [itex]R^2[/itex]. No, it is not a basis.[/B]
     
    Last edited by a moderator: Nov 14, 2015
  5. Nov 14, 2015 #4

    Krylov

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    I'm not sure that was the OP's intention, since
    but well... this makes things even more obvious.
     
  6. Nov 14, 2015 #5
    I actually meant for them to be column vectors, so v_1 = [1 2 -4] v_2 = [-4 3 6]
    I understand your explanation though. I just don't understand whether this is a basis or not if the vectors were column vectors. My guess is that it is, since it contains two pivots and is linearly independent, then it must be a basis for R_2. That last row is really throwing me off. I thought vectors in R_2 contain only two rows? Do the row sizes even matter?
     
  7. Nov 14, 2015 #6

    Mark44

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    The vectors you show here are vectors in ##\mathbb{R}^3##, so they definitely aren't in ##\mathbb{R}^2##, so can't span it, and can't be a basis for it. Just by observation I can see that these two vectors are linearly independent, which means that they are a basis for and span a two-dimensional subspace of ##\mathbb{R}^3##. Geometrically, this is a particular plane in three-dimensional space.
     
  8. Nov 14, 2015 #7
    Ah, it seems the trouble I am having is knowing the size of the subspaces. So, an ℝ3 subspace is not in three dimensions? I've always thought ℝ2 was a subspace in two dimensions, ℝ3 is a subspace in three dimensions, etc.. So the superscript k in ℝk only represents the rows in a column vector?

    so if I have the following vectors: x = x1 = [1 2 3 4 5 6]. x2 = [2 2 1 3 4 5], x3 = [1 2 9 9 2 1], x4 = [5 9 3 2 1 6]
    Written as column vectors [x1 x2 x3 x4] Could I say that the column space of A is in ℝ6? And the row space is in ℝ4? And it's null space is in ℝ4? And (if this set is linearly independent) its basis is in ℝ6?

    I'm just confused by how the book writes the ℝ as ℝn and ℝm. I know m represents the rows and n represents columns. In the Invertible Matrix Theorem, it states Col A = ℝn for an n x n matrix. So is this referring to the rows or the columns? The other confusing part the was mentioned in a theorem was that a column space of an m x n matrix A is a subspace of ℝm.

    Are column spaces of a matrix just its column vectors? Or is it the span of the column vector?
     
  9. Nov 14, 2015 #8

    Fredrik

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    For a set S to be a spanning set for a vector space V, or a basis for V, it must be a subset of V. You asked if two vectors in ##\mathbb R^3## span ##\mathbb R^2##. The set that contains those two vectors and nothing else, is a subspace of ##\mathbb R^3##, not ##\mathbb R^2##. Edit: I wrote that incorrectly. The sentence should say "spans a subspace", not "is a subspace".

    ##\mathbb R^3## has exactly one 0-dimensional subspace, infinitely many 1-dimensional subspaces, infinitely many 2-dimensional subspaces, and exactly one 3-dimensional subspace. The 0-dimensional subspace is {0}. The 3-dimensional subspace is ##\mathbb R^3##.

    ##\mathbb R^3## is a 3-dimensional vector space. ##\mathbb R^2## is a 2-dimensional vector space. Every vector space is a subspace of itself.

    A column vector has only one row. (Edit: Oops. I was pretty tired when I wrote that nonsense) The k tells you that the elements of the set are ordered n-tuples: ##(x_1,\dots,x_n)##. They can be written either as ##n\times 1## matrices or as ##1\times n## matrices. The former option has a number of advantages that make it more popular.

    I think it would be best if you study the basic definitions first (vector space, span, linearly independent, basis, subspace), and then ask your questions again.
     
    Last edited: Nov 15, 2015
  10. Nov 14, 2015 #9

    Mark44

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    There are many subspaces of the vector space ##\mathbb{R}^3##. The subspace that consists only of <0, 0, 0> is zero-dimensional. Any line through the origin will be a one-dimensional subspace. Any plane through the origin will be a two-dimensional subspace. ##\mathbb{R}^3## is considered to be a subspace of itself. What is common to all of these subspaces is that any vectors in any of them have three coordinates.
    ##\mathbb{R}^2## is a vector space. (It is also a subspace of itself.)
    ##\mathbb{R}^2## is a vector space. (It is also a subspace of itself.)
    Every vector in ##\mathbb{R}^k## will have k coordinates. Rows is not really the right term here.
    Yes, some subspace of ##\mathbb{R}^6##. If they are linearly dependent, the dimension of the subspace they span will be less.
    Yes
    Yes
    Yes and no, assuming you're still talking about ##x_1 \dots x_4##. When we talk about a basis, it's a basis for some subspace of whatever is the relavant vector space. We don't talk about the basis for a set of vectors.
    This doesn't make sense to me.
    The column space of a matrix is the subspace that is spanned by those vectors.
     
  11. Nov 14, 2015 #10

    Mark44

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    I disagree. A column vector has only on column. It could be thought of as an n x 1 matrix.
    Good advice...
     
  12. Nov 15, 2015 #11

    HallsofIvy

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    I frankly didn't notice that. But the column vectors are in [itex]R^3[/itex] and the OP specifically talked at [itex]R^2[/itex].
     
  13. Nov 15, 2015 #12

    Krylov

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    Yes, I understand. When I wrote my initial reply I read the OP's question the other way around (i.e. I read that he was talking about ##\mathbb{R}^3## instead of ##\mathbb{R}^2##). When I then read your reply, I realised that the answer to his question is even more obvious than I originally thought: the vectors are not even in the space.
     
  14. Nov 15, 2015 #13

    mfb

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    Just a technical detail:
    To get a subspace (instead of two isolated vectors), you have to take the span of that set of vectors.
     
  15. Nov 15, 2015 #14

    Fredrik

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    Thanks for catching those silly mistakes. This sort of thing happens a lot when I write a post a few hours past my bedtime.
     
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