My Math Forum Abstract Algebra Need Help

 Abstract Algebra Abstract Algebra Math Forum

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 Senior Member   Joined: Nov 2011 Posts: 100 Thanks: 0 Re: Abstract Algebra Need Help What have you tried so far? Are you stuck on a particular part, or just getting started?
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

 October 14th, 2012, 05:59 AM #4 Senior Member   Joined: Jul 2011 Posts: 227 Thanks: 0 Re: Abstract Algebra Need Help $M=\{\underline{w} | 0 = \underline{A}\underline{w}, \underline{w} \in \mathbb{R}^n \}$ To show: $(M,+)$ is a group. 1. First let's show that the operation $+$ is closed. Consider $\underline{w_1},\underline{w_2} \in M$ thus $\underline{A}\underline{w_1}=0$ and $\underline{A}\underline{w_2}=0$. $\underline{A}(\underline{w_1}+\underline{w_2})= \underline{A} \underline{w_1}+ \underline{A} \underline{w_2} =$ $0+0=0$ what means that $\underline{w_1}+\underline{w_2} \in M$ 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 $M$ 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.
 October 14th, 2012, 12:00 PM #5 Senior Member   Joined: Mar 2012 Posts: 294 Thanks: 88 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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