Linear Algebra Problems (Easy?)

  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. Dick

    Dick 25,913
    Science Advisor
    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. 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. Dick

    Dick 25,913
    Science Advisor
    Homework Helper

    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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