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anonymous

  • one year ago

@jvatsim ok so x1 = 3*x2 is the next problem so that means (3,1)^T (6,2)^T etc so nonempty here x1=3*x2 the next step alphaX so x = (3x2, x2)^T so alpha *x means (alpha*3*x2, alphax2)^T and that equals alpha*3*x2 = alpha *x2 equals alpha(3*x2 = x2) so it passes and on to step 3 x + y so here so i say x = (3x2,x2)^t y = (3y2, y2)^t we can still say 3x2+3y2 is = x2 + y2 ?? i think im missing something on this las step

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  1. anonymous
    • one year ago
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    @jtvatsim

  2. anonymous
    • one year ago
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    i misspelled your name above, sorry

  3. jtvatsim
    • one year ago
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    No problems. I'll be right back.

  4. jtvatsim
    • one year ago
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    First part looks good for nonempty.

  5. anonymous
    • one year ago
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    cool. the first step is the easy one

  6. jtvatsim
    • one year ago
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    Yep, you can always just show that the zero vector is in the set as a backup. Your way is fine though.

  7. jtvatsim
    • one year ago
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    Second part is a little convoluted, but I'll get back to that shortly, I think you are on the right track. :)

  8. anonymous
    • one year ago
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    how would you show the first step. i thought x1 was just 3 times x2 and since its in the x y plane its non empty

  9. jtvatsim
    • one year ago
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    First step I would just say (0,0) is in the set since 0 = 3*0. It's kind of "OK, duh..." but it works to prove the point.

  10. anonymous
    • one year ago
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    oh, wow ok yeah. makes sense

  11. jtvatsim
    • one year ago
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    For the second part, let's be clear first on what we will start with and what the goal is: START: x = (x1,x2) where x1 = 3x2. GOAL: alpha x = (alpha x1, alpha x2) where alpha x1 = 3(alpha x2). Notice the pattern.

  12. anonymous
    • one year ago
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    multiplying by alpha doesnt really change anything. is that what you are asking?

  13. anonymous
    • one year ago
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    so it is still 3x2 = x2 so it would pass, right?

  14. jtvatsim
    • one year ago
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    Yes, I think it is just a matter of writing style at this point. :) Here's how I would phrase the actual proof for clarity.

  15. jtvatsim
    • one year ago
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    Suppose we have x = (x1,x2). Then x1 = 3*x2. Now, alpha*x1 = alpha*3*x2, that is, alpha*x1 = 3*(alpha*x2). So, alpha*x = (alpha*x1, alpha*x2) is in the set. I would prefer to not convert x1 into 3*x2 in the proof. That turned me around a bit. But I see what you did. :)

  16. anonymous
    • one year ago
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    ok. so we have the same idea for step 2, just worked it differently?

  17. jtvatsim
    • one year ago
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    Wait... you say that "and that equals alpha*3*x2 = alpha *x2" but that isn't actually true.

  18. anonymous
    • one year ago
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    ok. if i go with what i was doing thats where i get a little confused for step 3 x + y for my way of working would be x = (3x2,x2) y = (3y2, y2) (3x2+3y2) would be 3(x2 + y2)...... should i just do what you did and make a note taht x1 = 3x2 and just use x1 in the proof?

  19. jtvatsim
    • one year ago
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    OK, let's hold off on part 2 for the time being. I want to try to do it your way for step 3. I think that might be more insightful.

  20. jtvatsim
    • one year ago
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    With your way, you have x = (3x2, x2) and y = (3y2, y2). OK, that's fine. What happens now is that we have no equation to work with. x1 = 3x2 is totally pointless right now since you already used it to convert. So, for x + y we need to show that x+y = (x1+y1, x2+y2) satisfies (x1+y1) = 3*(x2+y2). Is that actually true? Well you would say that x + y = (3x2 + 3y2, x2 + y2). And, yes, the first component 3x2 + 3y2 = 3*(x2 + y2) the second component. Thus, x + y is in the set. I have to say, it's a little unorthodox, but it seems to work.

  21. jtvatsim
    • one year ago
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    Your teacher is probably going to expect to see something with x1's in the proof, but I don't want to discount your way of doing it either. :)

  22. anonymous
    • one year ago
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    ok. i think your way would be better because its shorter. and i guess easier. i thought i had to use the 3x2 because the original problem said x1 = 3x2. so i thought we had to use it throughout

  23. jtvatsim
    • one year ago
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    Yeah, the equation is really just used as a "check" to make sure that the components work like they are supposed to.

  24. anonymous
    • one year ago
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    ok. im going to redo it and just make a note that x1 = 3x2 and use x1 in the proof

  25. anonymous
    • one year ago
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    i scrolled up and was reading and i think that is a lot less confusing than what i was trying

  26. jtvatsim
    • one year ago
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    Alright, I'd be happy to check it. It's always good to experiment with your own methods first so that you can sort out the process in your head. :)

  27. anonymous
    • one year ago
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    ok. ill be back in a few and post what i come up with

  28. anonymous
    • one year ago
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    just a quick question. we are saying \[\alpha x=0 \] right?

  29. anonymous
    • one year ago
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    for the second step

  30. jtvatsim
    • one year ago
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    We need to show that \[\alpha x\] satisfies \[\alpha x_1 = 3*(\alpha x_2)\]

  31. anonymous
    • one year ago
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    ok. so for step 2 (it applies to all problems depending on condition of x1 and x2) should i write \[\alpha x = 3*\alpha x2\]

  32. anonymous
    • one year ago
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    i think i answered my own question. yes

  33. anonymous
    • one year ago
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    before working it

  34. anonymous
    • one year ago
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    ok so for the third step (not doing what i originally did) would i again say x1=3x2 then X = (x1,x2) Y=(y1,y2) then X + Y = (x1+y1), (x2+y2) but that doesnt make sense. since i am trying to show x1 = 3x2. or would that actually be the proof because i cant think of a example that would counter that. (unless there is one and i cant think of it)

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