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Math Help - rank

  1. #1
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    rank

    in gaussian, is the number of vectors that form a basis of a solution set the rank or the dimension?

    i thought it would be the rank since for a vector to be a basis, it needs to be linearly independent...

    but im not very sure..

    thanks!
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  2. #2
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    Quote Originally Posted by alexandrabel90 View Post
    in gaussian, is the number of vectors that form a basis of a solution set the rank or the dimension?


    Say what?? What are you talking about, anyway?

    Tonio



    i thought it would be the rank since for a vector to be a basis, it needs to be linearly independent...

    but im not very sure..

    thanks!
    .
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  3. #3
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    sorry if i was not being clear.

    i was given 3 simultaneous equations and asked to find the number of vectors that form the basis for them..what i did was i did gaussian elimination..i was left with 2 leading ones and one free parameter. so from there, i concluded that the rank is 2 and the dimension is 1.

    now im wondering if the no of vectors that form B={ } is 2 vectors(the rank) or that one vector( that has the free parameter)
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  4. #4
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    Quote Originally Posted by alexandrabel90 View Post
    sorry if i was not being clear.

    i was given 3 simultaneous equations and asked to find the number of vectors that form the basis for them..what i did was i did gaussian elimination..i was left with 2 leading ones and one free parameter. so from there, i concluded that the rank is 2 and the dimension is 1.


    Well, at least this time it's a little clearer...but it seems to be that you were asked to find the dimension of the subspace spanned by those three equations (in the space of all equations with three unknowns over some given field), so that subspace's dimension = rank of the coefficients' matrix of the system is 2. Why do you write "dimension is 1"? Unless you mean, and I'm guessing since you said nothing, that 1 is the dimension of the kernel of that matrix = the matrix's null space.

    Then, the two non-zero lin. independent rows you get after applying Gauss method are a basis for that subspace=the span of those three equations).

    Tonio


    now im wondering if the no of vectors that form B={ } is 2 vectors(the rank) or that one vector( that has the free parameter)
    .
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  5. #5
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    nope. i thought dimension= number of free parameters = number of unknowns in the equation- rank = 3-2=1.

    however my notes also said that the answer was 2 (same as you)..

    did i mix up the definition of dimension?
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  6. #6
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    Quote Originally Posted by alexandrabel90 View Post
    nope. i thought dimension= number of free parameters = number of unknowns in the equation- rank = 3-2=1.

    however my notes also said that the answer was 2 (same as you)..

    did i mix up the definition of dimension?

    I'm afraid you did: for any matrix, the rank of its rows = the rank of its columns (and thus we can talk of "the matrix's rank" unambiguosly) = the dimension of the subspace spanned by the matrix's rows = the dimension of the subspace spanned by the matrix's columns.

    Tonio
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  7. #7
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    then what is the difference between rank and dimension?
    they seem like the same thing..

    thanks so much for clearing my confusion(:
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  8. #8
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    A matrix has a "rank", it does not have a "dimension". A vector space has a "dimension", it does not have a "rank". The rank of a matrix is the dimension of the vector space spanned by it columns (or rows) thought of as vectors. It is also the dimension of the subspace of all vectors of the form Av- that is the result of multiply A by any vector.
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