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Powers of the Matrix M^n

  1. Mar 10, 2009 #1
    1. The problem statement, all variables and given/known data
    I'm in 11th grade and I've been given the following in a series of problems:

    (2 0)
    (0 2)
    Calculate M^N for 1,2,3,4,5,10,20,50. Describe any patterns you observe. Generalize the pattern into an expression for the matrix M^n in terms of n.


    2. Relevant equations



    3. The attempt at a solution
    (2 0)
    (0 2)^2 =

    (4 0)
    (0 4)

    (2 0)
    (0 2)^3 =

    (8 0)
    (0 8)

    (2 0)
    (0 2)^4 =

    (16 0)
    (0 16)

    ((2 0)
    (0 2)^5 =

    (32 0)
    (0 32)

    (2 0)
    (0 2)^10 =

    (1024 0)
    (0 1024)

    (2 0)
    (0 2)^20 =

    (1048576 0)
    (0 1048576)

    It looks like to me that you can multiply the value in the prior matrix by 2 (for powers 1-5) to get the new value in the next one. For example:

    (2 0) (16 0)
    (0 2)^4 = (0 16) so multiply 16 by 2 and you have 32. You then know that the matrix to the power of 5 will look like this:
    (32 0)
    (0 32)

    Can someone help me find a rule in terms of n for M^n?
     
  2. jcsd
  3. Mar 10, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi agary12! Welcome to PF! :smile:
    Well, you're almost there …

    if MN =

    (kN 0)
    (0 kN)

    what is the rule for finding kN? :smile:
     
  4. Mar 10, 2009 #3
    I have a question about what you said,
    Is K being multiplied by N or is it just being included to show k is effected by N?

    using this particular example:

    (2 0)
    (0 2)^3 =

    (8 0)
    (0 8)

    What would the K value even be for the above value?
    To get 8 from 2 you have to put it to the 3rd power, but that doesn't give me any new information. You could also multiply it by 4, but there is no 4 in the problem.
     
  5. Mar 10, 2009 #4

    lanedance

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    i think tiny tim is saying try to find;

    k(n) = some function n...

    can you describe k(n)? (i think you pretty much described it in your last post...)
     
  6. Mar 10, 2009 #5

    tiny-tim

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    It's just an index :wink:

    (btw, try using the X2 and X2 tags just above the Reply box)
    ok, that's the N = 3 case …

    how about N = 4 … what's the pattern, and the mathematical rule of that pattern? :smile:
     
  7. Mar 10, 2009 #6

    rock.freak667

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    (2 0)
    (0 2)
    = 2* (1 0)
    (0 1)

    What does this matrix represent? Now what is any matrix mulitiplied by this matrix?
     
  8. Mar 10, 2009 #7

    HallsofIvy

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    He is suggesting that you look at the numbers 2, 4, 8, 16, 32, 64, 128, etc., which are what you get with n= 1, 2, 3, 4, 5, 6, 7, etc. What function of n are those?

    rock.freak667's suggestion, that you look at powers of [itex]2\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}[/itex] is also good.
     
  9. Mar 10, 2009 #8
    I'm still confused by what rock freak is trying to say. I understand that this:
    (2 0)
    (0 2)
    is equal to:
    2*(1 0)
    (0 1)
    but why is that even important? Are you suggesting I do something with this matrix?
    OH, I just typed it into my calculator and I think I may have found something. Basically this is just a
    (1 0)
    (0 1) matrix multiplied by 2
    The only thing that you are doing when you add an exponent is putting the 2 to a power, at which point it is distributed into the matrix right?

    [tex]\left\lfloor[/tex]
    K= 2, 4, 8, 16, 32, 64, 128
    N= 1, 2, 3, 4, 5, 6, 7

    Sorry but I'm not sure what you mena by "what function of n are those," are you saying that I need to find what is done to N to get K? In that case I am not sure. Nothing is being consistently multiplied by the N value to get K (N * X =/= k) since:
    2/1 = 2
    4/2 = 2
    8/3 = 2.66
    16/4 = 4
    32/5 = 6.4
    so there is no relationship found doing what I just did. I'm not sure what else I could do to N to get K.
     
    Last edited: Mar 10, 2009
  10. Mar 10, 2009 #9

    lanedance

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    there's nothing say its a linear relation with n, and in fact its clearly not

    you've said it a few times in the post so how about looking again at [tex]2^n [/tex] ?
     
    Last edited: Mar 10, 2009
  11. Mar 10, 2009 #10
    Why do you mean by looking at 2^N?

    Basically I've found that the 2 values other than zero in M^n are found by putting 2 to the power you are putting the overall matrix to.

    Therefore in this matrix
    (2 0) LaTeX Code:^5
    (0 2)

    I can find the new values by taking 2^5= 32
    The new matrix therefore is
    (32 0)
    (0 32)

    Is the rule then just 2^n to find the new values within the matrix? I'm not sure if this is what they are looking for.

    In response to tiny tim:

    See this is the problem, I understand that you are putting 2^3 to get 8 and that 2^4 is 16, but I'm not sure what the pattern is that they are looking for. They are simply powers of 2, but how can I say this mathematically? And what do you mean by the "mathematical rule of that pattern"?
     
  12. Mar 10, 2009 #11

    lanedance

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    write down M as a function of n. I think what they want is M(n)

    you've pretty much told us in previous posts what this is. M(n) is a multiple of the identity matrix, with scalar multiplier [tex] 2^n [/tex]

    this should be enough for what you're trying to do, but for a more general case you could write each specific element of M, to do this think about each element of M
    [tex]
    m_{ij}(n)
    [/tex], where i = row, j = column

    note m is diagonal, and multiply of the identity

    so what is [tex]
    m_{ij}(n)
    [/tex]
    when [tex]
    i = j
    [/tex]?
    and when [tex]
    i \neq j
    [/tex]?

    in short i think you've got everything you need, just have to pull it together...
     
  13. Mar 11, 2009 #12

    tiny-tim

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    Hi agary12! :smile:

    just got up :zzz: …

    oh i see!

    you've got it, but you think you haven't!

    how can you say mathematically "They are simply powers of 2"? …

    you say aN = 2N :wink:

    ok, so what is MN ? :smile:
     
  14. Mar 11, 2009 #13
    M^N equals:
    (2^N 0^N)
    (0^N 2^N)
    Is that it?

    Sorry lane dance, I'm not sure what your saying with the i and j subscripts. I haven't ever seen those before.
     
  15. Mar 11, 2009 #14

    tiny-tim

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    Yup! :biggrin:

    … except of course please write 0 not 0N

    (btw, rock.freak667 :smile: was saying that M = 2I (where I is the unit matrix), so MN = 2NIN = 2NI :wink:)
     
  16. Mar 11, 2009 #15

    lanedance

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    no worries agary12, look like you've got there, good worki

    If you find below confusing, don't worry about it for now, but thought I'll just add a bit for completeness:

    the i subscript relates to the row (horizontal line across matrix), so i = 1 is teh first row, i = 2 is the 2nd row & so on. Simialrly the j relates to the column (vertical line down matrix)

    So each element of teh matrix is identifed uniquely by a single i,j reference

    They don't have to be the letters i & j, its just a reference, could be any letter, but you see i & j quite often in books

    in terms of the identity matrix, it is often written
    [tex]
    \textbf{I}=
    [/tex]

    [tex]
    (1, 0)
    [/tex]
    [tex]
    (0, 1)
    [/tex]

    =
    [tex]
    (\delta_{11}, \delta_{12})
    [/tex]
    [tex]
    (\delta_{21}, \delta_{22})
    [/tex]

    where [tex]
    \delta_{ij} [/tex] is the kronecker delta defined by:
    [tex]
    \delta_{ij} = 1[/tex], if [tex]
    i = j [/tex]
    [tex]
    \delta_{ij} = 0[/tex], if [tex]
    i \neq j [/tex]

    in terms of your matrix it would look like
    [tex]
    \textbf{M}(n) =
    [/tex]

    [tex]
    (m_{11}, m_{12})
    [/tex]
    [tex]
    (m_{21}, m_{22})
    [/tex]

    =
    [tex]
    (2^n,0)
    [/tex]
    [tex]
    (0,2^n)
    [/tex]

    so you could write your formula for M as
    [tex]
    m_{ij} = 2^n\delta_{ij} [/tex]

    this is equivalent to

    M = 2n I
     
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