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Linear algebra- diagonalizability

  1. Dec 11, 2011 #1
    1. The problem statement, all variables and given/known data
    Compute A17.

    1 0 4
    A=0 1 2
    0 0 4


    2. Relevant equations

    A=PDP-1 (D is a diagonal matrix)

    3. The attempt at a solution

    I got eigenvalues 1 and 4, and corresponding eigenspaces u1=[1,0,0]T , u2=[0,1,1]T and u3=[4/3, 2/3 ,1]T.

    So, I computed P= (1/3) 3 0 4
    0 3 2
    0 0 3

    also, P-1= (1/3) 3 0 -4
    0 3 -2
    0 0 3.

    D= 1/3 1 0 0
    0 1 0
    0 0 4.


    and then A17= PD17P-1

    = 1/9 3 0 4[itex]\cdot[/itex]417 3 0 -4
    0 3 2[itex]\cdot[/itex]417 0 3 -2
    0 0 3[itex]\cdot[/itex]417 0 0 3

    = 1 0 4/3(417-1)
    0 1 2/3(417-1)
    0 0 417


    sorry for the matrix form here. I couldn't find any commands to do it correctly.

    Am I doing right here?
     
  2. jcsd
  3. Dec 11, 2011 #2
    oh... I cannot put spaces correctly :( everything here is 3x3 matrices.
     
  4. Dec 11, 2011 #3

    Deveno

    User Avatar
    Science Advisor

    to display matrices correctly use latex. for example,

    begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix} preceded by a slash,

    which gives the display:

    [tex]\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}[/tex]

    now if A = PDP-1, where D is diagonal,

    then A17 = PD17P-1

    and D17 is easy to compute, we just take the 17th power of the diagonal elements.

    it looks like you have calculated P correctly, but you wrote down the wrong eigenvector for u2, it should be [0,1,0]T. P-1, also looks ok to me.

    however, i get that D should just be:

    [tex]\begin{bmatrix}1&0&0\\0&1&0\\0&0&4\end{bmatrix}[/tex]

    which is going to throw off your values for A17
     
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