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Thread: Eigen valves question

  1. #1
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    Nov 2009
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    Eigen valves question

    Plz help me to find the answer for following question;

    A matrix

    $\displaystyle A= \begin{bmatrix}
    0 & 0 & 0 \\
    0 & \frac{29}{5} & \frac{12}{5} \\
    0 & \frac{12}{5} & \frac{36}{5}
    \end{bmatrix}
    $

    Find
    (a) Find Eigen valves A
    (b) Egen vectors A
    (c) obtain a matrix P such that $\displaystyle P^{-1} AP$ is a diagonal matrix
    (d) If $\displaystyle S=P^{-1} AP$, state the special property of each of $\displaystyle S $ and $\displaystyle S^{-1}$
    (e) Using the above result reduce the quadratic from Q(x) to form Q(y) where
    $\displaystyle Q(x)=16x_1^2 + \frac{29}{5} x_2^2 + \frac{36}{5} x_3^2 + \frac{24}{5} x_2 x_3$
    $\displaystyle Q(y)=16y_1^2 + +9y_2^2 + 4y_3^2$
    (f) obtain the relationship between $\displaystyle x^t = (x_1 ,x_2 ,x_3)^t$ and $\displaystyle y^t (y_1 ,y_2 ,y_3)^t$

    Now I solved part (a) as follows
    $\displaystyle A- \lambda I=\begin{bmatrix}
    16- \lambda & 0 & 0 \\
    0 & \frac{29}{5} - \lambda & \frac{12}{5} \\
    0 & \frac{12}{5} & \frac{36}{5} - \lambda
    \end{bmatrix}$
    the result gain $\displaystyle ( \lambda -16)( \lambda -4)( \lambda -9)$

    and

    part (b) solved as follows
    $\displaystyle
    \begin{pmatrix}
    16 & 0 & 0 \\
    0 & \frac{29}{5} & \frac{12}{5} \\
    0 & \frac{12}{5} & \frac{36}{5}
    \end{pmatrix}
    $
    x
    $\displaystyle \begin{pmatrix}
    a \\
    b \\
    c
    \end{pmatrix} = 16
    $

    then if $\displaystyle \lambda =4$
    $\displaystyle \begin{pmatrix}
    16 & 0 & 0 \\
    0 & \frac{29}{5} & \frac{12}{5} \\
    0 & \frac{12}{5} & \frac{36}{5}
    \end{pmatrix}
    $

    [tex]x
    $\displaystyle \begin{pmatrix}
    a \\
    b \\
    c
    \end{pmatrix} = 4$

    also I did for $\displaystyle \lambda =9$

    now I need help to find rest of parts in this question, also I need to know the above calculations are in correct or not!! if some are in incorrect plz. let me know the place and how I correct it
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  2. #2
    Junior Member
    Joined
    Jul 2009
    Posts
    69
    Quote Originally Posted by dhammikai View Post
    Plz help me to find the answer for following question;

    A matrix

    $\displaystyle A= \begin{bmatrix}
    0 & 0 & 0 \\
    0 & \frac{29}{5} & \frac{12}{5} \\
    0 & \frac{12}{5} & \frac{36}{5}
    \end{bmatrix}
    $

    Find
    (a) Find Eigen valves A
    (b) Egen vectors A
    (c) obtain a matrix P such that $\displaystyle P^{-1} AP$ is a diagonal matrix
    (d) If $\displaystyle S=P^{-1} AP$, state the special property of each of $\displaystyle S $ and $\displaystyle S^{-1}$
    (e) Using the above result reduce the quadratic from Q(x) to form Q(y) where
    $\displaystyle Q(x)=16x_1^2 + \frac{29}{5} x_2^2 + \frac{36}{5} x_3^2 + \frac{24}{5} x_2 x_3$
    $\displaystyle Q(y)=16y_1^2 + +9y_2^2 + 4y_3^2$
    (f) obtain the relationship between $\displaystyle x^t = (x_1 ,x_2 ,x_3)^t$ and $\displaystyle y^t (y_1 ,y_2 ,y_3)^t$

    Now I solved part (a) as follows
    $\displaystyle A- \lambda I=\begin{bmatrix}
    16- \lambda & 0 & 0 \\
    0 & \frac{29}{5} - \lambda & \frac{12}{5} \\
    0 & \frac{12}{5} & \frac{36}{5} - \lambda
    \end{bmatrix}$
    the result gain $\displaystyle ( \lambda -16)( \lambda -4)( \lambda -9)$

    and

    part (b) solved as follows
    $\displaystyle
    \begin{pmatrix}
    16 & 0 & 0 \\
    0 & \frac{29}{5} & \frac{12}{5} \\
    0 & \frac{12}{5} & \frac{36}{5}
    \end{pmatrix}
    $
    x
    $\displaystyle \begin{pmatrix}
    a \\
    b \\
    c
    \end{pmatrix} = 16
    $

    then if $\displaystyle \lambda =4$
    $\displaystyle \begin{pmatrix}
    16 & 0 & 0 \\
    0 & \frac{29}{5} & \frac{12}{5} \\
    0 & \frac{12}{5} & \frac{36}{5}
    \end{pmatrix}
    $

    [tex]x
    $\displaystyle \begin{pmatrix}
    a \\
    b \\
    c
    \end{pmatrix} = 4$

    also I did for $\displaystyle \lambda =9$

    now I need help to find rest of parts in this question, also I need to know the above calculations are in correct or not!! if some are in incorrect plz. let me know the place and how I correct it
    For matrix A, is row 1 column 1 suppose to be 16 or 0?

    for a, i didnt check, but it looks right?

    b. Eigenvectors should be a vector... not a number

    c. the diagonal values of S is the eigenvalues of A
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  3. #3
    Junior Member
    Joined
    Nov 2009
    Posts
    38
    Hi thanks for the reply and comments.
    Yes I got an error the matrix A should be
    $\displaystyle
    A= \begin{bmatrix}
    16 & 0 & 0 \\
    0 & \frac{29}{5} & \frac{12}{5} \\
    0 & \frac{12}{5} & \frac{36}{5}
    \end{bmatrix}
    $

    In part (b) the vector means if $\displaystyle \lambda =16$, then we have the vector of [a,b,c] is it true? likewise $\displaystyle \lambda =4$, and $\displaystyle \lambda =6$ also have vector [a,b,c] is it true?

    in part (c) you mean the eigen values of A is insert to the diagonal matrix?

    Plz explain me...
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  4. #4
    Junior Member
    Joined
    Jul 2009
    Posts
    69
    $\displaystyle A- \lambda I=\begin{bmatrix}
    16- \lambda & 0 & 0 \\
    0 & \frac{29}{5} - \lambda & \frac{12}{5} \\
    0 & \frac{12}{5} & \frac{36}{5} - \lambda
    \end{bmatrix}$
    the result gain $\displaystyle ( \lambda -16)( \lambda -4)( \lambda -9)$

    Assume ur eigenvalues are right, so $\displaystyle \lambda=16,4,9$

    For b, to find eigenvectors, sub each eigenvalues into the $\displaystyle A-\lambda I$ matrix and find the nullspace of the matrix, it should yield exactly 1 nullspace vector for each eigenvalue (for this example)

    c. yes, the diagonal entries of the diagonal matrix is the eigenvalues of A
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  5. #5
    Junior Member
    Joined
    Nov 2009
    Posts
    38
    Hi thnks for the reply and help.So now we complete part (a), I approch the part (b) as follows

    $\displaystyle \lambda =16$, the the mat will be
    $\displaystyle

    $
    $\displaystyle math]A= \begin{bmatrix}
    0 & 0 & 0 \\
    0 & \frac{51}{5} & \frac{12}{5} \\
    0 & \frac{12}{5} & \frac{44}{5}
    \end{bmatrix}
    $


    then according to equations $\displaystyle \frac{51}{5}y+ \frac{12}{5}z=0$ and $\displaystyle \frac{12}{5}y+ \frac{44}{5}z=0$, the results shold be x=0, y=0, z=o then the vector $\displaystyle v_1=(0,0,0)$

    If $\displaystyle \lambda=9$, then the matrix will be
    [/tex]$\displaystyle A= \begin{bmatrix}
    -70 & 0 & 0 \\
    0 & \frac{16}{5} & \frac{12}{5} \\
    0 & \frac{12}{5} & \frac{9}{5}
    \end{bmatrix}
    $
    Then according to equations
    $\displaystyle -7x=0$, $\displaystyle \frac{16}{5}y+ \frac{12}{5}z=0$ and $\displaystyle \frac{12}{5}y+\frac{9}{5}z=0$, results shouldbe x=0, y=3, z=-4, the vector is $\displaystyle v_2(0,3,4)$

    again $\displaystyle \lambda=4$, then the equations should be $\displaystyle -12x=0$, $\displaystyle \frac{-9}{5}y+\frac{12}{5}z=0$ and $\displaystyle \frac{12}{5}y-\frac{16}{5}z=0$. results are x=0, y=4, z=3

    plz help me the above procedure is correct or wrong..and if wrong let me know the right answer/way..
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