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...
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.
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 =...
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...
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...
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) +...
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...
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...
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)?
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...
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.....
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...
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...
do this, find this determinant:
\begin{vmatrix}x_1&y_1&ax_1+by_1\\x_2&y_2&ax_2+by_2\\x_3&y_3&ax_3+by_3 \end{vmatrix}
the results should be enlightening.
well, that's very nice, but:
in a group, the order of an element g is the smallest possible positive integer k such that gk = e.
what is gk? it's g*g*g*...*g (k times).
so when calculating order of an element, you don't need to look at "other elements".
in the group you are...
uh...i thought that that was what the word "implies" meant. i thought about using the word "forces" instead, making it clearer that the c's have to be all 0. i think this is just a semantical hitch, the idea is to set a linear combination to 0, and prove the coefficients must all be 0.
(to say...
two approaches to proving |P(A)| = 2|A|, for a finite set A.
approach 1: induction on |A|.
if |A| = 0, A is the empty set, in which case P(∅) = {∅} so |P(∅)| = 1 = 20.
assume that |P(A)| = 2|A|, whenever |A| < n.
now pick an element a in A. consider P(A-{a}). what sets are in P(A)...
i tend to think of it like this:
suppose we have a plane, which we will take to be R2.
now suppose we have a proper subspace of R2, which is of the form:
L = {a(x,y) : a in R}.
this is a line going through the origin and the point (x,y).
a coset is thus a set v + L, which is a...
U(ab) isn't always isomorphic to U(a) x U(b).
for example U(4) is cyclic of order 2, but U(2) x U(2) has but a single element: (1,1).
it IS true, if a and b are co-prime.
this is actually a consequence of the chinese remainder theorem:
if gcd(m,n) = 1, then [a]mn→([a]m,[a]n) is a...
we have to "reduce" somewhere...we can either put the "reduced form" in the definition of Q, or put the "reduced form" in the definition of H, but we can't neglect it all-together or our mapping may not be well-defined.
(the rationals are kinda funny that way: 1/2 isn't a rational number...
i always find it's best to go back to the definition:
a set of vectors {v1,v2,...vk} is linearly independent if:
c1v1+c2v2+...+ckvk implies:
c1 = c2 = ... ck = 0.
in this case, suppose that we have:
r1a + r2b = 0.
then we have r1a + r2b + r3c = 0
(by setting r3 = 0), and by...
you might want to prove some easier things first:
suppose 8|k, that is:
k = 2tm, where t ≥ 3.
then φ(k) = φ(2t)φ(m), so
U(k) ≅ U(2t) x U(m).
use this to show U(k) cannot be cyclic, as it has a non-cyclic subgroup.
now, consider even n and odd n for n2-1 separately.
the point of this problem is that 3 is PRIME. where are the factors of 3 in 2a, are they in the "a" part, or the "2" part?
is it even possible to put "half a factor of 3" somewhere in the "2 part", and the "other half" in the "a part"?
if not, then why not? part of what is means to be...