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Linear algebra: Find the span of a set

  1. Jan 21, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the span of [itex]U=\{2,\cos x,\sin x:x\in\mathbb{R}\}[/itex] ([itex]U[/itex] is the subset of a space of real functions) and [itex]V=\{(a,b,b,...,b),(b,a,b,...,b),...,(b,b,b,...,a): a,b\in \mathbb{R},V\subset \mathbb{R^n},n\in\mathbb{N}\}[/itex]

    2. Relevant equations
    - Span
    -Subset

    3. The attempt at a solution

    Objects in [itex]U[/itex] :[itex]2,\cos x,\sin x[/itex] are linearly independent, so they span [itex]\mathbb{R^3}[/itex].

    Let ,[itex]n=3\Rightarrow [V]= \begin{bmatrix}
    a & b & b \\
    b & a & b \\
    b & b & a \\
    \end{bmatrix}[/itex]

    [itex]rref[V]=\begin{bmatrix}
    1 & 0 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & 1 \\
    \end{bmatrix}\Rightarrow[/itex] vectors in [itex]V[/itex] span [itex]\mathbb{R^3}[/itex], if [itex]a,b\neq 0[/itex].

    But because [itex]V\subset\mathbb{R^n}\Rightarrow[/itex] vectors span [itex]\mathbb{R^{n-1}}[/itex].

    Is this correct?
     
  2. jcsd
  3. Jan 21, 2016 #2

    Mark44

    Staff: Mentor

    Yes.
    I don't know how valid your argument is, here. It's given that ##V \subset \mathbb{R}^n##. Can you extend this to a statement about an n-dimensional space instead of a 3-dimensional space?
    Why do you conclude that the vectors span ##\mathbb{R}^{n - 1}##?
     
  4. Jan 21, 2016 #3

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    They aren't even in ##\mathbb{R}^3##; they belong to a space of real functions.
     
  5. Jan 21, 2016 #4

    Mark44

    Staff: Mentor

    To clarify/correct my "Yes" in post #2, the functions span a three-dimensional space of functions, not ##\mathbb{R}^3##.
     
  6. Jan 22, 2016 #5
    Yes, these vectors are linearly independent.
    No, these vectors span a space isomorphic to [itex]\mathbb{R}^3[/itex]

    Regarding [itex]V[/itex]. If a=b, then all of those vectors are linearly dependent. If a=0, then the vectors will span a null space. If [itex]a\neq 0[/itex], then the system of vectors will be reduced to only one vector and so it would span [itex]\mathbb{R}^1\subset \mathbb{R}^n[/itex]
    If [itex]a\neq b[/itex], then all of the vectors are in fact linearly independent: one can construct an n x n matrix of said vectors and reduce it to a diagonal matrix.
     
    Last edited: Jan 22, 2016
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