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MORE linear algebra

  1. Jan 23, 2006 #1
    Let A be an m x n matrix with columns C1, C2, ... Cn. If rank A = n show taht [tex] \{ A^{T}C_{1}, A^{T}C_{2}, ... , A^{T}C_{n} /} [/tex] is a basis of Rn.

    ok [tex] \mbox{rank} A^{T} = n [/tex]
    the columns of A are rows of A transpose
    im not sure how to proceed though...
    a column times itself with [tex] C_{1}^2 + C_{2} C_{1} + ... + C_{n}C_{1} [/tex] for the first term of [tex] A^{T} C_{1} [/tex] is the rank maintained through this multiplication? What justifies that?

    help is greatly appreciated!!!

    thank you!
     
  2. jcsd
  3. Jan 23, 2006 #2

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    What (product of) matrix (-x + ces) has the ATCi as its columns? And what property does an n x n matrix have to have for its rows to form a basis of R^n?
     
    Last edited: Jan 23, 2006
  4. Jan 23, 2006 #3
    what do u mean (-x + ces) ?
    arent the rows of a square matrix A linearly independant if they form a basis for Rn?
     
  5. Jan 23, 2006 #4

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    Sorry, it was supposed to be like "giraffe(s)," but that's not as easy when the word ends in an x. Anyway, right, they form a basis if the matrix containing them as columns has rank n.
     
  6. Jan 23, 2006 #5
    ok...
    im still not sure to do with the -x + Ci part.
     
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