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October 7th, 2012, 08:08 AM   #1
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Abstract Algebra Need Help

Hello, I have 2 Problems
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October 8th, 2012, 08:17 AM   #2
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Re: Abstract Algebra Need Help

What have you tried so far? Are you stuck on a particular part, or just getting started?
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October 13th, 2012, 09:00 AM   #3
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Re: Abstract Algebra Need Help

Quote:
Originally Posted by watson
What have you tried so far? Are you stuck on a particular part, or just getting started?
The first task I knew how to solve independently.

d+(d+d)=d+d=d
d+(d+a)=d+a=a
d+(d+b)=d+b=a

(d+d)+d=d+d=d
(d+d)+a=d+a=a
(d+d)+b=d+b=b


Please help me with the second, I do not understand it at all, read some books, but did not understand ...
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October 14th, 2012, 05:59 AM   #4
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Re: Abstract Algebra Need Help


To show: is a group.
1. First let's show that the operation is closed. Consider thus and .

what means that
2. The operation is associative. Check it by yourself, it's almost the same as 1.
3. Find the neutral element for this operation. Can you do that?
4. Every element of has an invers element. Can you check that?

Furthermore the group is abelian because the vector addition is commutative.

For the second question. I don't exactly know what they mean with a distinction of all the constant matrices. What you have to do here is solving a linear system.
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October 14th, 2012, 12:00 PM   #5
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Re: Abstract Algebra Need Help

for problem 3, we must show two things for the first part:

a) if u,w are in M, then u+w is also in M.

b) if w is in M, then -w is in M.

that is all we need to show (M,+) is a group. we get associativity of + "for free", since vector addition in R^n is associative. and from a) & b) we get that:

for any w in M: -w and w+(-w) are in M, and in R^n, w+(-w) = 0, so that the 0-vector of R^n is an identity for M.

we are given that Au = 0, and Aw = 0 (where these are products of an mxn matrix with an nx1 matrix on the LHS, and the 0 is the mx1 0-vector of R^m).

thus A(u+w) = Au + Aw (since matrix multiplication is linear)

= 0 + 0 = 0, which shows (a).

also A(-w) = A((-1)(w)) = (-1)(Aw) = (-1)0 = -0 = 0 (again, all 0's here are mx1 0-vectors, NOT the real number 0. the -1 however, is the real number -1 (a scalar)).

this shows (b).

it's hard to know exactly what is being asked in the second part.

if m = n, and rank(A) = n, then the only solution is {0}, the 0-vector of R^n. rank(A) = n if and only if A is an invertible matrix. if rank(A) < n, then A maps some non-zero vector u in R^n to the 0-vector in R^n, so A(cu) = c(Au) = c0 = 0, so cu is also in M, and we get an infinite number of elements in M (one for every real number c, at least).

if m > n, it is pretty much the same as above, except that rank(A) = n does not mean A has an inverse. we still have {0} as the only element of M if rank(A) = n, and infinite elements in M if rank(A) < n.

if m < n, then rank(A) ? m < n, so we ALWAYS have an infinite number of elements in M.

note that no matter WHAT m and n are, the 0-vector of R^n is always an element of M: (M.+) is a group and groups cannot be empty (the identity element lies in any group).
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