Recent content by Deveno

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    Can someone look at my proof?

    that is a fair point.. professors usually appreciate it when students don't "get ahead of themselves" and use theorems/concepts that haven't been covered yet. that said, there's no harm in "looking ahead" and understanding that what you have done can be viewed an alternate way. it's like...
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    Orthogonal Basis and Inner Products

    you're fine up to here. at this point, stop thinking about the "inner products", and start thinking about which linear combination of the Ai, A has to be.
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    Mod or quotient remainder theorem (QRT)

    my guess is, you are learning some number theory. my second guess is, you are more comfortable with writing: a = b + kn than: a = b (mod n). what i am "doing with the subscripts" is this: [a]n = b, where a = b + kn, and 0 ≤ b < n. the reason for the brackets is that [a]n =...
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    Can someone look at my proof?

    rather than use "not an integer" arguments, it would be cleaner to stay totally within the integers. that is, if 4k2+ 5 = 4t then 5 = 4(t - k2) → 4|5, impossible. similarly, if 4k2 + 4k + 9 = 4t then 9 = 4(t - k2 + k) → 4|9, also impossible. ******* alternate proof # 1: if 4|n2 + 5, then...
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    Extract h

    silly me. you're right of course.
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    Mod or quotient remainder theorem (QRT)

    i think this isn't even the right approach (you could get there from here, but it's the long way around). suppose we write [a]k, for the equivalence class (or residue class, i.e., the remainder upon division by k) of the integer a. then once can DEFINE: ([a]k)([b]k) = [ab]k. in...
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    Vectors: finding the angle between two vectors using dot product method when you don

    what SammyS means is that the dot product is bilinear, it is linear in each variable: if a,b,c are vectors, and r is a scalar: a.(b+c) = a.b + a.c (a+b).c = a.c + b.c a.(rb) = r(a.b) (ra).b = r(a.b) also, a.b = b.a (the dot product is symmetric). thus (a+5b).(2a-3b) = 2(a.a) +...
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    Matrix Theory problem

    your previous hint was: "multiply either equation by Q". i count 5 "=" in the OP's post, so it is unclear to me which two of them you mean. i suppose you mean: 1) P-1AP = B 2) Q-1AQ = B note that "multiply by Q" is not unambiguously defined, since Mat(n,F) is a non-commutative monoid...
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    Extract h

    yes, but every term contains a h, so the non-zero solutions are the roots of a quadratic, which is somewhat easier.
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    How long can I expect to work on Spivak?

    if memory serves me, Spivak was written to cover a full year course. trying to cover it "all" in 16 weeks might be asking a bit much. the first 4 chapters will probably be fairly easy going, although some of the exercises may make you stop and think for a bit. chapters 7 & 8 are...
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    Matrix Theory problem

    i have an alternate approach: note that B = P-1AP is the same as: A = PBP-1. also, note that if R-1AR = A, then AR = RA. what happens if you evaluate (QP-1)-1A(QP-1)?
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    Question involving a linearly independent set of vectors

    i understand your point Mark44, and it's well-taken. as Ray pointed out, clarification doesn't hurt, as the "goal" of understanding linear independence is not to be able to state an impeccable definition of it, but rather, to be able to actually determine if sets are linearly independent or not...
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    Could a saint please guide me work through my research on AFFINE SETS AND MAPPINGS

    where does "f" come from? my reasoning goes like this: 1/2 and 1/2 sum to 1, so (1/2)x + (1/2)y is an affine combination, that is: (x+y)/2 is in M. now, use part (iii) to conclude that.....
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    Cosets and Vector Spaces Question

    suppose u + W = v + W. then (u - v) + W = (v - v) + W = 0 + W = W (i simply subtracted v + W from "both sides", using the fact that -(v + W) = (-1)(v + W) = (-1)v + W = -v + W). by definition of a subspace, u - v is also in U, if both u,v are (subspaces are closed under vector addtion and...
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    Finite abelian group into sequence of subgroups

    there are a couple of "easier" cases you might want to look into first: |G| = pq, p,q distinct primes. |G| = pk, p a prime. the second case is "harder", although you may have proved both of these already if you have covered the sylow theorems. by the way, any group of prime order is...
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