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Re: Vector Space Problem I Need Help

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  • beata_stehlikova
    While -1 = 1 there are three suspaces: [(1,0)] = {(1,0), (0,0)} [(0,1)] = {(0,1), (0,0)} [(1,1)] = {(1,1), (0,0)} But you are true that it is a
    Message 1 of 4236 , Sep 1 4:19 AM
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      While -1 = 1 there are three suspaces:<br>[(1,0)]
      = {(1,0), (0,0)}<br>[(0,1)] = {(0,1),
      (0,0)}<br>[(1,1)] = {(1,1), (0,0)}<br>But you are true that it is a
      counterexample to the expression [n:k].<br><br>I think that the
      number of subspaces should depend not only on n,k but
      also on p.<br><br>To my previous message<br>It is
      better to consider a set of k independent vectors
      instead of the set which contains k ind. vector. <br>Then
      you get the reduced triangular matrix with non zero
      rows and you don't need to think of deleting zero
      rows.<br>I realised it after posting.<br><br>Beata
    • sum_what_ever
      can t sleep (yawn), i spent around 3 hr.s on your your original post here. it seems this implicit function solution set is the semi-stable modular eliptical
      Message 4236 of 4236 , Apr 25, 2011
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        can't sleep (yawn), i spent around 3 hr.s on your your original post here. it seems this implicit function solution set is the semi-stable modular eliptical curves A. Wiles gave in his proof of fermat's last theorem. Your function is way simpler. have to get some sleep before i got to go to work, but the proof of this is around 4 to 5 pages as you said. i'll have to look at it more this evening. also, what's the scoop of you being the source for the plot of tron legacy? we should all get royalties! heheh.
        --- In mathforfun@yahoogroups.com, jeshields_98 wrote:
        >
        > I hit the wrong button...<br><br>It should be = c^n<br>not + c^n<br><br>very VEEERRRRRYYYYY important<br><br>sorry
        >
        I
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