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Homework Help: Linear Algebra Problems (Easy?)

  1. Feb 17, 2010 #1
    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. jcsd
  3. Feb 17, 2010 #2


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    Homework Helper

    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?
  4. Feb 17, 2010 #3
    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
    -Ax2 -Ax2

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


    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.
  5. Feb 18, 2010 #4


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    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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