
#1
Feb1710, 08:44 PM

P: 7

This stuff is confusing. I don't know if it's hard or not, I just have a feeling I don't really know what I'm doing.
1. The problem statement, all variables and given/known data 1. Show that the equation Ax=b has a unique solution if and only if the solution to Ax=0 is x=0. 2. Let A be an m x p matrix, and let B be a p x n matrix. Show that the range of A is contained in the range of AB. Show that the kernel of B is contained in the kernel of AB. Is the reverse inclusion true in either case? 2. Relevant equations 1. Ax=b; Ax=0; x=0 2. See below. 3. The attempt at a solution 1. I really have no idea what to do. 2. A=[v1 ... vp]; B=[w1 ... wn] AB=[Aw1 ... Awn] im(A)=c1v1 + ... + cpvp im(AB)=c1Aw1 + ... + cnAwn=A(c1w1 + ... + cnwn) That's all I have. This is probably not even close to what I'm supposed to be doing. Please help! 



#2
Feb1710, 09:17 PM

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P: 25,171

No, you don't know what you are doing. You need practice. Let's just start with the first one. Suppose Ax=b has two different solutions. Ax1=b and Ax2=b with x1 not equal to x2. If you subtract those two equations what does that tell you about solutions to Ax=0?




#3
Feb1710, 11:24 PM

P: 7

Okay, here's my attempt at solving the first.
Suppose Ax=b has two solutions, x=x1 and x=x2, where x1=/=x2. Ax1=b Ax2=b Ax1=Ax2 Ax2 Ax2 Ax1Ax2=0 A(x1x2)=0 Now let's assume x1=x2; this condition implies that Ax=b has a unique solution. A(x1x2)=0 A0=0 We thus see that the equation Ax=b has a unique solution if and only if the unique solution to Ax=0 is x=0. Is this extraneous or wrong? Also, what are you thoughts on the second problem? Any suggestions for practice? Thanks so much. 



#4
Feb1810, 08:15 AM

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Linear Algebra Problems (Easy?)
You've got the right idea. The phrasing could use some work. You assumed x1 and x2 are two different solutions, you conclude A(x1x2)=0.
Ok but now don't "assume x1=x2". You've got x1x2=/=0. That's another solution to Ax=0 besides x=0. So you've got "if the solution to Ax=b isn't unique then the solution to Ax=0 isn't unique". That's the same as saying "the solution to Ax=0 is unique implies the solution to Ax=b is unique", right? Now since the question says "if and only if" you should do the opposite direction as well. For the second one, think of the matrices as linear maps. A maps R^p to R^m. AB maps R^n to R^m by going through R^p. Look up the definition of 'range' and think about it. 


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