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Matrix^n without eigenvalues

  1. Jan 20, 2010 #1
    Oh it gives me headache... been thinking on this problem for a while, and don't even know where to begin! Could any one give me a hint at least?? :(

    Let A be (3x3) matrix : [ 4 -2 2; 2 4 -4; 1 1 0] and u (vector) = [1 3 2].
    a) Verify that Au = 2u
    I got this one without a problem.

    b) Without forming A^5, calculate the vector A^5*u.
    This is where I get stuck.. I've tried to search, but keep coming up with some equations that involve eigenvalues (which I havent studied yet..). So, im assuming that i don't have to use any of eigenvalues.. Is there any other way?
    I tried to replace A matrix with [a b c; d...] values, and take first 3 powers of that, but its way too hard to keep track of everything..

    So, any hint?? :(
  2. jcsd
  3. Jan 20, 2010 #2


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    You know that Au = 2u.

    Now what is A2u = A (A u)?

    Note that numbers commute with matrices, so for example A(2u) = 2(Au).

    Actually you are using eigenvalues here: when Av = c v for some (non-trivial) vector v and some number c, then we say that v is an eigenvector for the matrix A, with corresponding eigenvalue c.
  4. Jan 20, 2010 #3
    Oh man! Thats simple! Didn't even think about relating problem a with problem b! :rolleyes: Thanks a ton! :biggrin:
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