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Inductive proof for matrix multiplication problem

  1. Sep 15, 2009 #1
    1. The problem statement, all variables and given/known data

    Compute

    [1 1]^n
    [ 1]

    This is what my book has. I am assuming the blank means 0 (please tell me if this is not the case).

    2. Relevant equations

    None really (matrix multiplication)

    3. The attempt at a solution

    I did n = 1, 2, 3, 4 and found a patten. It looks like:

    [1 1]^n
    [ 1]
    =
    [1 n]
    [ 1]

    But how do I show that? Obviously showing it for a couple cases doesn't mean it's actually true. Inductively, I can show the n = 2 case is true. but I don't really know how to show the n+1 case is true other than write

    [1 1]^n+1
    [ 1]
    =
    [1 n+1]
    [ 1]

    but that seems a little unsatisfactory or like I'm skipping steps or something. How do you do it?
     
  2. jcsd
  3. Sep 15, 2009 #2

    Mark44

    Staff: Mentor

    Assume that the statement is true for an exponent of n. Then use this to show that the (n + 1)st power of your matrix produces the right value.

    The blank means 0, I'm pretty sure.
     
  4. Sep 15, 2009 #3
    Right. But how do you do that?

    Letting n be the natural numbers and assuming the following statement is true

    [1 1]^n
    [ 1]
    =
    [1 n]
    [ 1]

    Thus,

    [1 1]^n+1
    [ 1]
    =
    [1 n+1]
    [ 1]


    I don't really know how to show it beyond that - is there some abstract way to write out the intermediate steps? As it is, it seems somewhat unsatisfactory.
     
  5. Sep 16, 2009 #4

    Mark44

    Staff: Mentor

    I'm putting the 0 in the lower left corner since it looks silly to leave it out.

    [1 1]^(n+1)
    [0 1]

    =
    [1 1]^n [1 1]
    [0 1] [0 1]

    = ??
    This is where you use what you have assumed.
     
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