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anonymous

  • one year ago

not sure where to start for this proof. Determine whether the following sets form subspaces of R^2 {(x1,x2)^T | x1+x2=0}

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  1. jtvatsim
    • one year ago
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    Do you know the conditions required to have a subspace?

  2. anonymous
    • one year ago
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    it has to be nonempty alpha x is in the subset for any alpha x+y is in the subset that is what my book says. but not sure how to apply that to this

  3. jtvatsim
    • one year ago
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    OK, there are three properties as your book suggests. Another equivalent version says: 1) the zero vector must be in the subset (this ensures nonemptiness, as well as other useful things) 2) constant multiples must be in the subset (alpha x, where alpha is a scalar and x is a vector) 3) vector sums must be in the subset (x + y, where x and y are both vectors).

  4. jtvatsim
    • one year ago
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    The trick, as you mention, is figuring out how to apply it. :)

  5. jtvatsim
    • one year ago
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    We are dealing with the subset {(x1,x2)^T | x1 + x2 = 0} correct? Are you clear on what all the notation means here?

  6. anonymous
    • one year ago
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    i think one of the problems is that calculus 1 was the only pre req for linear but trying to read the book i think its assumed you have had discrete or some other math proof based class. at least thats what i get from talling to other students

  7. anonymous
    • one year ago
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    talking*

  8. jtvatsim
    • one year ago
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    Yes, you are probably correct... I imagine it is quite a bit more challenging in that case... :) It is easy enough once you get the basics. I'll try to walk you through this one if you think that will be most helpful.

  9. anonymous
    • one year ago
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    anything would be. im lost and tried googling stuff but its just more proofs and symbols that i then have to try and google what those symboles mean

  10. jtvatsim
    • one year ago
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    OK. Let's start with step one. Identify what all the important concepts, symbols, and definitions are.

  11. anonymous
    • one year ago
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    ok. i know that x1 and x2 are vectors in R^2

  12. anonymous
    • one year ago
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    and im assuming R^2 is the x y plane

  13. anonymous
    • one year ago
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    and there sum is 0

  14. jtvatsim
    • one year ago
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    Good, so far. There is a minor issue however.

  15. freckles
    • one year ago
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    well I think (x1,x2)^T is a vector in R^2 not x1 and x2

  16. anonymous
    • one year ago
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    thats about as far as i can go

  17. anonymous
    • one year ago
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    what issue?

  18. jtvatsim
    • one year ago
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    That's a good start. As @freckles mentioned, we need to be careful to distinguish that x1 and x2 are not vectors separately.

  19. jtvatsim
    • one year ago
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    The whole expression is (x1,x2) which is a vector. x1 is the x-component, and x2 is the y-component. These individual components sum to 0.

  20. anonymous
    • one year ago
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    so its a single vector like x = (x1 x2)

  21. jtvatsim
    • one year ago
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    Correct

  22. jtvatsim
    • one year ago
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    It is useful to think of an actual example just so we have something concrete in mind. So some examples of vectors that satisfy this are (1,-1) since 1 + (-1) = 0, (-2, 2) since (-2) + 2 = 0. and so on. Make sense so far?

  23. anonymous
    • one year ago
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    yes. i guess "abstractly" x1 = negative x2 so that they sum to 0

  24. jtvatsim
    • one year ago
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    Exactly! In fact, that is what the expression x1 + x2 = 0 implies. Solving for x1 gives as you say, x1 = -x2.

  25. jtvatsim
    • one year ago
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    Alright, so we still need to decide whether this set of vectors is in fact a subspace.

  26. jtvatsim
    • one year ago
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    Let's start with the first condition: 1) Does the zero vector live in this set?

  27. anonymous
    • one year ago
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    i would say yes because 0 + (-0) is still 0. is that the right way to think of it?

  28. jtvatsim
    • one year ago
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    You are 100% correct!

  29. jtvatsim
    • one year ago
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    (0,0) satisfies the condition that x1 + x2 = 0, since 0 + 0 = 0. Good, our set passes condition 1.

  30. jtvatsim
    • one year ago
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    Now, to check condition 2: If c is some multiple and v is in the set, is cv in the set?

  31. jtvatsim
    • one year ago
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    Sorry, I guess I should say If alpha is some multiple and x is in the set, is alpha x in the set?

  32. jtvatsim
    • one year ago
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    It's the same thing either way.

  33. jtvatsim
    • one year ago
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    So, this is a little weird, but here goes.

  34. anonymous
    • one year ago
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    so alphax1 + alphax2 = 0 so alpha is any number alpha = 100 x1 = 5 so x2 would = -5 so someting like 100(5) + 100(-5) = 0 because 500 + -500 = 0

  35. jtvatsim
    • one year ago
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    Great, that's the spirit of the proof! Now we just make the argument as general as possible.

  36. anonymous
    • one year ago
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    thats where i run into trouble.

  37. anonymous
    • one year ago
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    becasue i asked him if i can give an example on a test that is coming up and he said no because one example might be right but there is posibly infinite examples and one could be false and some other stuff i didnt quite catch

  38. jtvatsim
    • one year ago
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    Here is how to phrase the general proof. Look it over and if you have any questions, we will spend some time unpacking it. "Suppose we have x = (x1,x2) in the set. So, x1 + x2 = 0. Now, look at alpha x. alpha x = (alpha x1, alpha x2). Notice that alpha x1 + alpha x2 = alpha (x1 + x2). But we know that x1 + x2 = 0! So, alpha x1 + alpha x2 = alpha (x1 + x2) = 0 as we wanted."

  39. jtvatsim
    • one year ago
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    Notice that we need to start with x. We do not know if alpha x1 + alpha x2 actually equals 0. This is what we are trying to show... Instead we start with x and show that alpha x works too. With a couple of clever algebraic tricks we pull it off.

  40. anonymous
    • one year ago
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    ok i think i get it, one question though, might be dumb but where you (and my book) say constant multiples must be in the subset. lets say the subset is 1 2 3 4 5 does that mean alpha can only be 1 2 3 4 5 and not 6 or anything higher or am i completely wrong on that

  41. anonymous
    • one year ago
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    but what you did is only for the second condition, right?

  42. jtvatsim
    • one year ago
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    This is only the second condition, yes. And your question is a good one. Let me think how to answer it.

  43. jtvatsim
    • one year ago
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    alpha can be as large or as small as it wants. It must work for any alpha. So, in your example, you give a very small subset. {1, 2, 3, 4, 5}. This is actually NOT a subspace since alpha = 10 breaks it. In fact 10*1 = 10 which is not in the set.

  44. jtvatsim
    • one year ago
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    Does that answer your question?

  45. anonymous
    • one year ago
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    ok, yes it does

  46. jtvatsim
    • one year ago
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    Great. One last note on condition 2. We have to start with x1 + x2 = 0 NOT alpha x1 + alpha x2 = 0 since the set only talks about x1 + x2 = 0 in the definition.

  47. jtvatsim
    • one year ago
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    That was a minor difference in your proof versus my general proof.

  48. jtvatsim
    • one year ago
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    Just wanted to be sure you picked up on that. :)

  49. jtvatsim
    • one year ago
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    So where are we so far? We have shown that condition 1 and 2 both work. All that remains is condition 3. If it works, then we have a subspace. If it fails, we do not have a subspace.

  50. anonymous
    • one year ago
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    i scrolled up and read what you said the three conditinos are and number 3 says someting about 2 vectors but dont be only have 1 vector x ?

  51. jtvatsim
    • one year ago
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    Good point. We are allowed to assume we have a second vector y for condition 3's proof.

  52. jtvatsim
    • one year ago
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    After all, with condition 2 we just showed that this set probably has an infinite number of vectors (every multiple works).

  53. anonymous
    • one year ago
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    so would the second vector consist of 0's ?

  54. jtvatsim
    • one year ago
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    Well, we don't exactly know. It is safer to say that the second vector y = (y1,y2) will just act like the other vectors in this set. That is, y1 + y2 = 0.

  55. anonymous
    • one year ago
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    because anything added to 0 is still whatever that number was. so 5 + 0 = 5?

  56. anonymous
    • one year ago
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    is that right or wrong way to think of it?

  57. anonymous
    • one year ago
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    ok so then again y1 = -y2

  58. jtvatsim
    • one year ago
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    I see your point. However, your example is dealing with numbers rather than vectors.

  59. jtvatsim
    • one year ago
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    Yes, that is the safer assumption.

  60. jtvatsim
    • one year ago
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    But hold your thought, you might need just a slight change in thinking to see what we are getting at.

  61. jtvatsim
    • one year ago
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    So, how will we know if x + y is in the set?

  62. jtvatsim
    • one year ago
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    We need to show that its components add to 0.

  63. anonymous
    • one year ago
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    if the vectors of x = -y ?

  64. jtvatsim
    • one year ago
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    Yes, I think you could do it that way.

  65. anonymous
    • one year ago
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    is that not the right way? or was there something else?

  66. jtvatsim
    • one year ago
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    Ah... wait. We aren't saying that x + y = 0. We want to say that (first components) + (second components) = 0.

  67. jtvatsim
    • one year ago
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    Here's what I mean.

  68. jtvatsim
    • one year ago
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    First, we need to know what x + y looks like. x + y = (x1,x2) + (y1,y2) = (x1+y1,x2+y2)

  69. jtvatsim
    • one year ago
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    Then, we need to find a way to show that x1+y1 + x2+y2 = 0. That will be the trick.

  70. jtvatsim
    • one year ago
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    Does that make sense? Any questions on that part?

  71. anonymous
    • one year ago
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    i think so. wouldt you just move x2 and y2 over? so x1 + x2 = -x2 - y2 ??

  72. anonymous
    • one year ago
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    actually i think thats wrong

  73. jtvatsim
    • one year ago
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    You could do that. But that won't be a proof. We need to start with the most basic assumption and build up to what we want. So here we go (*long inhale*)... :)

  74. jtvatsim
    • one year ago
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    "Suppose we have two vectors x and y in the set."

  75. jtvatsim
    • one year ago
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    "Then, x = (x1,x2) so x1 + x2 = 0 and y = (y1,y2) so y1 + y2 = 0. This is true since they are in the set."

  76. jtvatsim
    • one year ago
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    "Now, I'd like to think about x + y. x + y looks like this: x + y = (x1,x2) + (y1,y2) = (x1+y1,x2+y2)."

  77. jtvatsim
    • one year ago
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    "Now, do these components sum to 0? I notice that (x1 + y1) + (x2 + y2) = x1 + y1 + x2 + y2 = (x1 + x2) + (y1 + y2)"

  78. jtvatsim
    • one year ago
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    "But, wait, I just said before that x1 + x2 = 0 and y1 + y2 = 0. So, the components do sum to 0 after all! In fact, (x1 + y1) + (x2 + y2) = (x1 + x2) + (y1 + y2) = 0 + 0 = 0 as I wanted."

  79. jtvatsim
    • one year ago
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    Do you follow? There are little nuances here and there, but this is really the same format you follow for any "subspace proof". :)

  80. jtvatsim
    • one year ago
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    I'll make a summary of the highlights so you can bring all these scattered ideas together. :)

  81. anonymous
    • one year ago
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    ok. im still reading everything you just typed and comprehending, might take a minute

  82. jtvatsim
    • one year ago
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    No worries, take your time, these things don't happen in a flash. It took me a lot of practice to get comfortable. :)

  83. jtvatsim
    • one year ago
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    This is what a finished proof might look like (you might find it helpful to see the big picture all at once): {(x1,x2)^T | x1 + x2 = 0} is a subspace of R^2. Proof: A set is a subspace if it is 1) nonempty, 2) cx is in the set for any scalar c, and 3) if x + y is in the set. 1) (0,0) is in the set since 0 + 0 = 0. So, the set is nonempty. 2) Suppose x is in the set and c is some scalar. Then x = (x1,x2) and x1 + x2 = 0. Think about cx. cx = (cx1, cx2). Now cx1 + cx2 = c(x1 + x2) = c(0) = 0. So cx is in the set. 3) Suppose x and y are in the set. Then x = (x1,x2) and x1 + x2 = 0, also y = (y1,y2) and y1 + y2 = 0. Think about x + y. x + y = (x1,x2) + (y1,y2) = (x1+y1,x2+y2). Now (x1 + y1) + (x2 + y2) = (x1 + x2) + (y1 + y2) = 0 + 0 = 0. So x + y is in the set. Since all three conditions are met, the set is a subspace.

  84. anonymous
    • one year ago
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    that was both kind of morerinvolved than i was thinking but not that bad once you explained it. im still reading. but it is making sense

  85. jtvatsim
    • one year ago
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    Yes, proofs are quite picky, but the format is almost always the same.

  86. jtvatsim
    • one year ago
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    The most frustrating thing is not being able to just start with the goal. We cannot say: "Suppose alpha x1 + alpha x2 = 0..." and go from there. We must say: "Suppose x is in the set. Then x1 + x2 = 0..." and somehow manipulate this to get to "alpha x1 + alpha x2 = 0". It's a subtle but crucial difference in direction.

  87. anonymous
    • one year ago
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    ok. i have some similar problens to work on. like x1*x2 = 0 and a few others but im going to try and apply what you explained to them. i might hit you up if i get stuck but im going to try them first

  88. jtvatsim
    • one year ago
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    Sounds good. Remember that not all sets will be subspaces. If you an find a way to break any of the three rules you are allowed to say "No way, this is not a subspace."

  89. jtvatsim
    • one year ago
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    Also a "Pro Tip," subspaces are linear, so if you have multiplication of variables (like x1*x2) you will probably not have a subspace.

  90. anonymous
    • one year ago
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    so is there a different way to work it?

  91. jtvatsim
    • one year ago
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    If you don't have a subspace, just write down the example that breaks it. Are you working on the x1*x2 = 0 one?

  92. anonymous
    • one year ago
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    yes

  93. jtvatsim
    • one year ago
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    OK, I'm looking at it now.

  94. jtvatsim
    • one year ago
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    Well, let's just go through the list. Does (0,0) work?

  95. anonymous
    • one year ago
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    so is x = (1,0)

  96. anonymous
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    or x = (0,1)

  97. jtvatsim
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    Yep, all those are in the set so far.

  98. anonymous
    • one year ago
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    well yes 0 *0 =0

  99. jtvatsim
    • one year ago
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    Good, so no way are we going to break rule 1. Pass.

  100. anonymous
    • one year ago
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    ok so as long as one of them is 0 them it passes

  101. jtvatsim
    • one year ago
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    Yes, I like that summary. Much simpler than what I was going to say. :)

  102. jtvatsim
    • one year ago
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    Alright, but look at rule 3 for a second. x + y. Well (1,0) is in the set, and (0,1) is in the set. But what about if we add these together? (1,0) + (0,1) = (1,1)... Uh oh...

  103. jtvatsim
    • one year ago
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    Do you see the problem here?

  104. anonymous
    • one year ago
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    yes. it would be 1 and not 0

  105. jtvatsim
    • one year ago
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    And that is a BIG problem. So, how would we write this as a proof? Many ways to do it, but here's one.

  106. anonymous
    • one year ago
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    ok. before that, is there a reason you went straight to 3 and skipped 2??

  107. jtvatsim
    • one year ago
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    Well, I happened to realize that rule 3 would break immediately and didn't want to waste brain energy on rule 2. After all, if only one breaks, we know instantly that it is not a subspace.

  108. jtvatsim
    • one year ago
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    Actually, if you think about it, rule 2 works since we have at least one 0 in the component, and multiplying 0 by any number won't do anything. c(x1,0) = (cx1,0) or c(0,x2) = (0,cx2)

  109. jtvatsim
    • one year ago
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    So Rule 1 and Rule 2 pass. But Rule 3 fails. Once we know that a single rule fails, we are allowed to stop and declare the result.

  110. jtvatsim
    • one year ago
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    Does that make sense? :)

  111. anonymous
    • one year ago
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    yes. im still working on writing a "proof" but yeah step 3 fails so we would say it is not in the subspace of R^2 because it failed step 3

  112. jtvatsim
    • one year ago
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    Here's an example proof that I came up with to help: {(x1,x2) : x1*x2 = 0} is not a subspace. Proof: If the set were a subspace x + y should be in the set whenever x and y are in the set. But, look at (1,0) and (0,1). These are both in the set since 1*0 = 0 and 0*1 = 0. But, (1,0) + (0,1) = (1,1). 1*1 is not 0, so the sum is not in the set. Therefore, the set is not a subspace.

  113. jtvatsim
    • one year ago
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    A proof is really just the "short version" of the long conversations we are having here. There are very few people who can think like a proof. Most of us have to stumble through examples and making sense of the question before we then go back through our thought process and figure out what was important.

  114. anonymous
    • one year ago
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    ok. so the proof is using variables like x1, x2, y1, y2 etc for the "abstract" big picture and we can include numerical examples if they would fail one of the 3 steps

  115. jtvatsim
    • one year ago
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    Yes. It is much easier to prove something false, than to prove something true. When proving something is true, we must use the abstract big picture to account for every single possible, imaginable, thing. When proving something is false, we can use simple numerical examples since if a statement fails even once it is false.

  116. jtvatsim
    • one year ago
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    This is why it is truly amazing that anything could be a subspace, because the requirements are very tight and narrow. It is mind blowing that those three rules would hold forever no matter how many examples you tried.

  117. anonymous
    • one year ago
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    is kind of cool. someone a lot smarter than me must have had to do a ton of problems to figure that out. i have more problems with variations of x1 and x2 but im going to close this question and try and work them. and if i get stuck ill @ you or something. might even do it just to check me if you dont mind

  118. anonymous
    • one year ago
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    but thanks for all your help

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