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Math Help - Matrix question

  1. #1
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    Matrix question

    A = \left( {\begin{array}{rr}<br />
   {-10} & {-18}  \\<br />
   9 & 17  \\<br /> <br />
 \end{array} } \right)

    a) Find Y such that Y^{-1}AY is diagonal.

    b) Find C such that C^3 = A


    For a I get Y = \left( {\begin{array}{rr}<br />
   2 & 1  \\<br />
   -1 & -1  \\<br /> <br />
 \end{array} } \right)

    I can't figure out b. Does it involve the Cayley-Hamilton theorem at all? I've only used that for solving C=A^3 not this way round. Thanks!
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  2. #2
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    Hello buddy

    Quote Originally Posted by Thomas154321 View Post
    A = \left( {\begin{array}{rr}<br />
   {-10} & {-18}  \\<br />
   9 & 17  \\<br /> <br />
 \end{array} } \right)

    a) Find Y such that Y^{-1}AY is diagonal.

    b) Find C such that C^3 = A


    For a I get Y = \left( {\begin{array}{rr}<br />
   2 & 1  \\<br />
   -1 & -1  \\<br /> <br />
 \end{array} } \right)
    This is correct, in this case it is

    Y^{-1}AY = diag(-1,8)

    b) I don't know, sorry
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  3. #3
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    Thanks that's reassuring.

    I've also found another question I can't do:

    Give an example of a square matrix over \mathbb{C} which is not similar to a diagonal matrix, and explain why this is the case.


    No clue here. I know the two matrices are similar if they represent the same linear map with respect to two different bases, but other than that I don't know.
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  4. #4
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    Joined
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    Hi, regarding b):

    we know that
    <br />
Y^{-1}AY = \begin{bmatrix}-1 & 0 \\ 0 & 8 \end{bmatrix}.<br />

    And hence A = YDY^{-1}.

    Define C = \begin{bmatrix}-1 & 0 \\ 0 & 2 \end{bmatrix}.

    Then C^3 = D.

    Also,

    <br />
YCY^{-1})^3 = YCY^{-1}YCY^{-1}YCY^{-1} = YC^3Y^{-1} = YDY^{-1} = A,<br />


    so
    YCY^{-1}
    is the matrix we are looking for.

    If I work this out, I get

    <br />
YCY^{-1} = \begin{bmatrix} -4 & -6 \\ 3 & 5 \end{bmatrix}.<br />

    Best,

    ZD
    Last edited by ZeroDivisor; April 26th 2009 at 06:12 AM. Reason: Latex formatting
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  5. #5
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    Quote Originally Posted by Thomas154321 View Post
    A = \left( {\begin{array}{rr}<br />
{-10} & {-18} \\<br />
9 & 17 \\<br /> <br />
\end{array} } \right)

    a) Find Y such that Y^{-1}AY is diagonal.

    b) Find C such that C^3 = A


    For a I get Y = \left( {\begin{array}{rr}<br />
2 & 1 \\<br />
-1 & -1 \\<br /> <br />
\end{array} } \right)

    I can't figure out b. Does it involve the Cayley-Hamilton theorem at all? I've only used that for solving C=A^3 not this way round. Thanks!
    Let M = diag(-1, 8). Then M = Y^{-1} A Y \Rightarrow A = Y M Y^{-1}.

    Therefore the answer to b) is C = Y N Y^{-1} where N = diag(-1, 2) since (Y N Y^{-1})^3 = Y N^3 Y^{-1} = Y M Y^{-1}.
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