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Linear Algebra Problems (Easy?)

by gasaway.ryan
Tags: algebra, linear
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gasaway.ryan
#1
Feb17-10, 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!
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Dick
#2
Feb17-10, 09:17 PM
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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?
gasaway.ryan
#3
Feb17-10, 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
Ax1-Ax2=0
A(x1-x2)=0

Now let's assume x1=x2; this condition implies that Ax=b has a unique solution.

A(x1-x2)=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.

Dick
#4
Feb18-10, 08:15 AM
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P: 25,228
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(x1-x2)=0.
Ok but now don't "assume x1=x2". You've got x1-x2=/=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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