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Matrix powers

  1. Oct 25, 2004 #1
    hi guys, today i was confonted with this problem in grade 11 high school math.

    consider the matrix M=(2 0)
    (0 2)

    calculate M^n for n = 2,3,4,5,10,20,50

    and find a general expression for the matrix M^n in terms of n.

    this problem has troubled me a lot, and no matter how hard i tried, i couldnt find a solution...could someone help me out here?
    thanks guys
  2. jcsd
  3. Oct 26, 2004 #2


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    Homework Helper

    Find M^2, M^3, M^4 and M^5 by hand. What do you get? do you see a pattern?
  4. Oct 27, 2004 #3
    yeah...the first and the last element are 2^the power given.

    but how do i put this into a equation?
    thanks again for all your help!
  5. Dec 2, 2007 #4
    M^n = M*2^(n-1) is the equation for this particular matrix only. The equation differs for others.
  6. Dec 3, 2007 #5


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    Since you expressed M as
    (2 0)
    (0 2)
    what's wrong with Mn as
    (2n 0)
    (0 2n) ?
  7. Feb 9, 2008 #6
    I have the same question and am just wondering what your final general formula was... the (2^n 0)
    (0 2^n) one only works when there are zeros on the diagonal and I have a question where the matrix is (3 1)
    (1 3)
    -I don't understand the outcome answers when it is squared :
    (10 6)
    (6 10) -is there a general formula to use? What are matrices like this called?
  8. Feb 9, 2008 #7
    Any matrix with unique eigenvectors and eigen values can be diagonalized:

    A=V D V^-1

    Where D is a diagonal matrix of the eign values
    The ith column of V is the eign vector which corresponds to the eign value on the diagnal of the ith column of D.

    A^2=V D V^-1 V D V^-1=V D (V^-1 V) D V^-1=
    V D I D V^-1=V D D V^-1=V D^2 V^-1

    In general

    A^N=V D^N V^-1

    Your above expression is simple enough that just by doing regular multiplication you might be able to see the pattern without applying the above theory.
  9. Feb 10, 2008 #8
    The matrix [itex]M[/itex] can be written as [itex]M=2\,I_2[/itex], where [itex]I_2[/itex] is the unit [itex]2\times2[/itex] matrix. Thus [itex]M^2=2\,I_2\cdot 2\,I_2=2^2\, I_2^2=2^2\,I_2[/itex]. Try the same thing for [itex]M^3,\,M^4,\dots[/itex] If you want to prove that [itex]M^n=2^n\,I_2[/itex] use induction.
  10. Feb 10, 2008 #9
    Just noticed that this thread is almost 4 months old! :rofl:
  11. Feb 10, 2008 #10
    Thanks for the help-is there a simpler way to prove that-I've never heard of the stuff you are referring to...this is for grade 11 math. Thanks again though!
  12. Feb 10, 2008 #11


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    Well, I didn't learn it until I was in college!
  13. Mar 16, 2008 #12
    I got the exact same problem, and am pretty sure that there is some relatively 11th grade adequate way to approach this. Maybe something with step 3, where (k+1 k-1) this matrix applies to both the example (2 0) and (3 1)
    _________________(k-1 k+1) __________________________________(0 2) ___(1 3)

    I really can't find a general equation for the last one in terms of k and n...
    Last edited: Mar 17, 2008
  14. Mar 17, 2008 #13

    After some serious work on my last post i came up with this, feedback please!
    This is matrix Mk raised to the nth power.

    M[tex]^{N}_{K}[/tex] = [tex]\left([(k+1) +(k - 1)\sum^{n}_{x=1} k^{x}] [(k-1) +(k - 1)\sum^{n}_{x=1} k^{x}] \right)[/tex]
    [tex]\left([(k - 1) +(k - 1)\sum^{n}_{x=1} k^{x}] [(k + 1) +(k - 1)\sum^{n}_{x=1} k^{x}] \right)[/tex]

    (The thumbnail is a lot clearer)

    Attached Files:

    Last edited: Mar 17, 2008
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