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Linear Independce Outise R^n

  1. Oct 14, 2012 #1
    To text whether n vectors in R^n are linearly independent, you put those vectors in a matrix and take its determinant.

    How can this be generalized beyond vectors in R^n-- say to the space of matrices in R^(mxn)?
     
  2. jcsd
  3. Oct 15, 2012 #2

    chiro

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    Hey schaefera.

    In this new space do you have mxn vectors in mxn space? If so you do exactly the same thing except your matrix is (mxn)x(mxn).

    If you want to check whether any set of vectors are linearly dependent (below the dimension of the space), simply put the vectors in a matrix and do a reduced-row echelon reduction on the matrix and see what it's rank is. The rank will give you the number of linearly independent vectors for that set that you entered in.
     
  4. Oct 16, 2012 #3
    How about for the space of continuous functions? Polynomials? Does the method ever break down?
     
  5. Oct 16, 2012 #4
    The question of linear independence of a finite amount of vectors can be thought of as asking about solutions to the equation: [tex] c_1\mathbf{v}_1+ c_2\mathbf{v}_2+...+c_n\mathbf{v}_n= \mathbf{0}[/tex] where [itex]\mathbf{v}_i \in V[/itex] are the vectors you are testing for linear independence and [itex] c_i \in F[/itex] are scalars in your field [itex]F[/itex]. The vectors [itex]\mathbf{v}_i[/itex] will be linearly independent if and only if all the scalars, [itex] c_1,c_2,...,c_n[/itex] are zero. Said another way: [tex]c_1\mathbf{v}_1+ c_2\mathbf{v}_2+...+c_n\mathbf{v}_n= \mathbf{0} \Rightarrow c_1,c_2,...,c_n=0[/tex] So to test a set of vectors for linear independence you set up the above equation and check to see if all the scalars [itex]c_i[/itex] are zero. How exactly you go about checking that is more or less dependent on what vector space you're dealing with. As for the case of infinitely many vectors I'm not 100% sure, so I won't comment.
     
  6. Oct 17, 2012 #5

    HallsofIvy

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    What Gamble93 gives is the usual definition of "linear independence". Requiring that a matrix having non-zero determinant is a specific property.
     
  7. Oct 17, 2012 #6
    I realized after my post that I should have included that a non zero determinant implying linear independence is a specific case that applies to [itex]\mathbb{R}^n[/itex] but I was late for class so it slipped my mind. Excuse my ignorance.
     
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