Study HN: Linear Algebra 1, fortnight 2

Re:Study HN: Linear Algebra 1, fortnight 2 by gruseom on Wednesday October 19, 2011 @02:38AM (#3129023)

That's exactly the sort of interpretation I was looking for. I don't remember how to represent geometrical transformations as matrices, but that's really just a technical detail; one can understand your solution without it. That being said, I still don't understand your solution. Are you saying that any such pair of transformations as you describe has the desired anticommutative property, except for the exceptions? I don't see either part of that - the general claim or the exceptions.

My solution exploits the fact that in CD, C combines the rows of D, while in DC, C combines the columns of D. Here's the rest:

Yrg P or gur vqragvgl zngevk rkprcg jvgu -1 nf vgf obggbz-evtug ryrzrag. Gura PQ vf Q jvgu vgf obggbz ebj artngrq. Fvzvyneyl, QP vf Q jvgu vgf evtug pbyhza artngrq naq -QP vf Q jvgu vgf yrsg pbyhza artngrq. Fb gb trg PQ = -QP, jr whfg arrq gb svaq n Q fhpu gung artngvat vgf obggbz ebj unf gur fnzr rssrpg nf artngvat vgf yrsg pbyhza. Sbe 2k2 zngevprf, guvf vf nal Q jvgu mreb va gur gbc-yrsg naq obggbz-evtug cbfvgvbaf.

Rot13 don't work so good for specifying matrices!

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Re:Study HN: Linear Algebra 1, fortnight 2 by nialo on Wednesday October 19, 2011 @01:08PM (#3131040)

I apparently entered my test cases wrong in lots of interesting ways. I think I've actually found essentially the same solution you have, it turns out I had the exceptions and the actual solution backwards:

n avargl qrterr ebgngvba naq n ersyrpgvba nobhg, nf sne nf V pna gryy nal, yvar guebhtu gur bevtva frrzf gb tvir gur evtug erfhyg.

Are we sure this is the correct meaning for "A = -B", where A and B are matrices? Some things I've read seem to be saying that that should be equivalent to AB = identity matrix, rather than Av = -1(Bv), where v is some vector.

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Re:Study HN: Linear Algebra 1, fortnight 2 by gruseom on Thursday October 20, 2011 @02:04AM (#3133650)

It's the correct meaning, as indicated by the solution on p. 430. You're thinking not of -B but of the inverse of B, which is denoted B^-1 ("B to the power of -1") and when multiplied by B gives I.

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