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Diagonalizable matrix

  1. May 4, 2008 #1
    1. The problem statement, all variables and given/known data

    Use diagonalization to find A^9 where A = [0 8;2 0]

    3. The attempt at a solution

    I perform (A-λI) to find characteristic equation and hence eigenvalues of 4 and -4. These yield eigenvectors of [2;1] and [2;-1] respectively.

    Now I know [0 8;2 0][2 2;1 -1] = [2 2;1 1][4 0;0 -4]
    Rearranging gives [0 8;2 0] = [2 2;1 -1][4 0;0 -4][2 2;1 -1]^-1

    Now to raise to the power of 9 gives, [0 8;2 0]^9 = [2 2;1 -1][4 0;0 -4]^9[2 2;1 -1]^-1

    Now I know that any matrix multiplied by its inverse will give the identity matrix, hence this leaves me with [0 8;2 0]^9 = [4 0;0 -4]^9

    Therefore by my working, A^9 = [4^9 0; 0 (-4)^9]

    The answer suggests [0 2*4^9; .5*4^9 0]

    Can someone see where I went wrong?
     
  2. jcsd
  3. May 4, 2008 #2

    Defennder

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    You're still correct here. Let the matrices on the RHS be labeled P,D^9,P^-1

    This is your mistake. You're multiplying PD^9P^-1, not D^9P(P^-1). In matrix multiplication the order matters. You can't randomly switch the order. Just multiply the matrices out normally and you'll get the answer.
     
  4. May 4, 2008 #3
    Ah, so if I have matrices ABC and I wish to perform A*B*C
    I must do (A*B)*C.

    hmmm, when I did it for a previous question that asked to perform A^6 (where A was the same matrix) I multiplied P and P^-1 to give the inverse and that yielded the correct answer. Just a lucky once off I suppose?
     
  5. May 4, 2008 #4

    Defennder

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    Matrix multiplication is associative, meaning ABC = (AB)C = A(BC). But it is not commutative, meaning that ABC not necessarily equals ACB.

    I guess you were lucky that time. I didn't try it out and I can't see your working so I can't tell.
     
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