Results 1 to 5 of 5

Thread: Matrix question

  1. #1
    Member
    Joined
    Sep 2006
    Posts
    77

    Matrix question

    $\displaystyle A = \left( {\begin{array}{rr}
    {-10} & {-18} \\
    9 & 17 \\

    \end{array} } \right)$

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

    b) Find C such that $\displaystyle C^3 = A$


    For a I get $\displaystyle Y = \left( {\begin{array}{rr}
    2 & 1 \\
    -1 & -1 \\

    \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 $\displaystyle C=A^3$ not this way round. Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2008
    Posts
    461
    Hello buddy

    Quote Originally Posted by Thomas154321 View Post
    $\displaystyle A = \left( {\begin{array}{rr}
    {-10} & {-18} \\
    9 & 17 \\

    \end{array} } \right)$

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

    b) Find C such that $\displaystyle C^3 = A$


    For a I get $\displaystyle Y = \left( {\begin{array}{rr}
    2 & 1 \\
    -1 & -1 \\

    \end{array} } \right)$
    This is correct, in this case it is

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

    b) I don't know, sorry
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2006
    Posts
    77
    Thanks that's reassuring.

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

    Give an example of a square matrix over $\displaystyle \mathbb{C}$ 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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Apr 2009
    Posts
    24
    Hi, regarding b):

    we know that
    $\displaystyle
    Y^{-1}AY = \begin{bmatrix}-1 & 0 \\ 0 & 8 \end{bmatrix}.
    $

    And hence $\displaystyle A = YDY^{-1}.$

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

    Then $\displaystyle C^3 = D.$

    Also,

    $\displaystyle
    YCY^{-1})^3 = YCY^{-1}YCY^{-1}YCY^{-1} = YC^3Y^{-1} = YDY^{-1} = A,
    $


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

    If I work this out, I get

    $\displaystyle
    YCY^{-1} = \begin{bmatrix} -4 & -6 \\ 3 & 5 \end{bmatrix}.
    $

    Best,

    ZD
    Last edited by ZeroDivisor; Apr 26th 2009 at 06:12 AM. Reason: Latex formatting
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    Quote Originally Posted by Thomas154321 View Post
    $\displaystyle A = \left( {\begin{array}{rr}
    {-10} & {-18} \\
    9 & 17 \\

    \end{array} } \right)$

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

    b) Find C such that $\displaystyle C^3 = A$


    For a I get $\displaystyle Y = \left( {\begin{array}{rr}
    2 & 1 \\
    -1 & -1 \\

    \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 $\displaystyle C=A^3$ not this way round. Thanks!
    Let M = diag(-1, 8). Then $\displaystyle M = Y^{-1} A Y \Rightarrow A = Y M Y^{-1}$.

    Therefore the answer to b) is $\displaystyle C = Y N Y^{-1}$ where N = diag(-1, 2) since $\displaystyle (Y N Y^{-1})^3 = Y N^3 Y^{-1} = Y M Y^{-1}$.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Matrix Question
    Posted in the Algebra Forum
    Replies: 0
    Last Post: Feb 3rd 2010, 11:12 AM
  2. a question on 4x4 matrix
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: Aug 6th 2009, 11:06 PM
  3. Matrix question! can someone help please!!!
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Jul 17th 2009, 04:08 AM
  4. Matrix Question
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Jul 1st 2009, 03:31 AM
  5. Matrix Question
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Jun 10th 2008, 08:58 PM

Search Tags


/mathhelpforum @mathhelpforum