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R^n and R^m

  1. Oct 7, 2012 #1
    In this theorem it states " Let A be a m x n matrix. That is For each vector b in R^m, the column Ax=b has a solution..." Why do they say that bεR^m? Is that because b is a mx1 column matrix where it has m rows making it belong to R^m?
     
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  3. Oct 7, 2012 #2

    HallsofIvy

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    If A is an "m by n matrix" then it has n columns and m rows. That means that to multiply it by vector v, written as a column matrix, v must have n components, and then the product with be a vector, again written as a column matrix, will have m components. That is, A is from Rn to Rm.
     
  4. Oct 7, 2012 #3
    so what your saying is that (mxn)(mx1)=(mx1).
    and that means that bεR^m?
     
    Last edited by a moderator: Oct 7, 2012
  5. Oct 7, 2012 #4

    HallsofIvy

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    I'm not sure I know what you mean by "(mxn)(mx1)= (mx1)". I thought we were talking about matrices, not numbers. What I said before was that a matrix with n columns and m rows, multiplied by a matrix with 1 column and n rows, gives a matrix with 1 column and n rows.

    If I understand you notation, that would be "(mxn)(nx1)= (mx1)".
     
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