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Math Help - Eigen valves question

  1. #1
    Junior Member
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    Eigen valves question

    Plz help me to find the answer for following question;

    A matrix

    A= \begin{bmatrix}<br />
  0      & 0 & 0      \\<br />
0 & \frac{29}{5} & \frac{12}{5} \\ <br />
  0      & \frac{12}{5} & \frac{36}{5}<br />
\end{bmatrix}<br />

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

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

    and

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

    then if \lambda =4
    \begin{pmatrix}<br />
  16 & 0 & 0 \\<br />
  0 & \frac{29}{5} & \frac{12}{5} \\<br />
0 & \frac{12}{5} & \frac{36}{5} <br />
\end{pmatrix}<br />

    [tex]x
    \begin{pmatrix}<br />
a \\<br />
  b \\<br />
c <br />
\end{pmatrix} = 4

    also I did for \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

    A= \begin{bmatrix}<br />
  0      & 0 & 0      \\<br />
0 & \frac{29}{5} & \frac{12}{5} \\ <br />
  0      & \frac{12}{5} & \frac{36}{5}<br />
\end{bmatrix}<br />

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

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

    and

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

    then if \lambda =4
    \begin{pmatrix}<br />
  16 & 0 & 0 \\<br />
  0 & \frac{29}{5} & \frac{12}{5} \\<br />
0 & \frac{12}{5} & \frac{36}{5} <br />
\end{pmatrix}<br />

    [tex]x
    \begin{pmatrix}<br />
a \\<br />
  b \\<br />
c <br />
\end{pmatrix} = 4

    also I did for \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
    <br />
A= \begin{bmatrix}<br />
16 & 0 & 0 \\<br />
0 & \frac{29}{5} & \frac{12}{5} \\<br />
0 & \frac{12}{5} & \frac{36}{5}<br />
\end{bmatrix}<br />

    In part (b) the vector means if \lambda =16, then we have the vector of [a,b,c] is it true? likewise \lambda =4, and \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
    A- \lambda I=\begin{bmatrix}<br />
  16- \lambda      & 0 & 0      \\<br />
0 & \frac{29}{5} - \lambda & \frac{12}{5} \\ <br />
  0      & \frac{12}{5} & \frac{36}{5} - \lambda<br />
\end{bmatrix}
    the result gain ( \lambda -16)( \lambda -4)( \lambda -9)

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

    For b, to find eigenvectors, sub each eigenvalues into the  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

    \lambda =16, the the mat will be
    <br /> <br />
math]A= \begin{bmatrix}<br />
0 & 0 & 0 \\<br />
0 & \frac{51}{5} & \frac{12}{5} \\<br />
0 & \frac{12}{5} & \frac{44}{5}<br />
\end{bmatrix}<br />

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

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

    again \lambda=4, then the equations should be -12x=0, \frac{-9}{5}y+\frac{12}{5}z=0 and \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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