Page 1 of 2 12 LastLast
Results 1 to 15 of 19

Math Help - Finding diagonalize matrix

  1. #1
    Newbie
    Joined
    May 2010
    Posts
    16

    Finding diagonalize matrix

    I need help finding an orthogonal matrix P which diagonalize the matrix M6

    Matrix M6:=
    17 16 16
    16 17 16
    16 16 17

    I think that the eigenvalues of the matrix are 1 and 49/33 but i'm not sure.
    Also i don't know how to continue from this.

    Please help me!

    /Maria
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    I would agree with the 1 as an eigenvalue. However, the 49/33 is incorrect. To find the orthogonal matrix P, you'll need to orthonormalize the eigenvectors via Gram-Schmidt. What do you get when you do that?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2010
    Posts
    16
    The problem is that i dont understand how to get the eigenvectors. If 1 is the eigenvalue of every row in the matrix i get

    A - (lambda)*I gives the following matrix

    16 16 16
    16 16 16
    16 16 16

    Row reducing this matrix gives me

    1 1 1
    0 0 0
    0 0 0

    From here I dont know how to find the eigenvectors since both x_2 and x_3 are free

    edit

    After doing some more reading i realised that get the basis

    x_2 =[-1,1,0]
    x_3=[-1,0,1]

    But i cant figure out the basis for x_1.

    If i understand this correctyl i need 3 eigenvectors since the first matrix is 3x3.
    Last edited by maria88; November 3rd 2010 at 05:21 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Well, there may be two eigenvectors. Set x_{2}=s and x_{3}=t, and solve for all three components in terms of those two parameters, just like you normally do in Gaussian elimination with back substitution. What do you get?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    May 2010
    Posts
    16
    I don't understand, should I put t and s as answers for x_2 and x_3in the original matrix as such:

    \begin{bmatrix}17&16&16&0\\16&17&16&s\\16&16&17&s\  end{bmatrix}

    and then do a gaussian elimination? If so i get the following equations:
    x_{1}+\dfrac{16}{17}x_{2}+\dfrac{16}{17}x_{3}}=0

    x_{2}+\dfrac{16}{17}x_{3}=\dfrac{17s}{33}}

    x_{3}=\dfrac{33t}{49}-\dfrac{16s}{49}}

    From these equations I am not sure how to move on to find the eigenvectors.



    My first a attempts before using s and t gave me the following equations:


    x=\begin{bmatrix}x{_1}\\x{_2}\\x{_3}\end{bmatrix}= x_{2}\begin{bmatrix}-1\\x1\\x0\end{bmatrix}x_{3}\begin{bmatrix}-1\\0\\1\end{bmatrix}

    This would give me the matrix \begin{bmatrix}-1&-1\\1&0\\0&1\end{bmatrix}

    Normalizling this gave me: \begin{bmatrix}-1&\dfrac{-1}{2}\\1&\dfrac{-1}{2}}\\0&1\end{bmatrix}

    Am I going the right way in any of these cases?

    / Maria
    Follow Math Help Forum on Facebook and Google+

  6. #6
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    No, to find the eigenvectors x_{\lambda} corresponding to a particular eigenvalue \lambda (you always have to find an eigenvector relative to a particular eigenvalue), you solve the linear system (A-\lambda I)x_{\lambda}=0 for x_{\lambda}. What do you get when you do that? (This is the system in post # 3.)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    May 2010
    Posts
    16
    I'm not sure I understand what you are saying, sorry english isn't my first language.

    Should I substract the identity matrix multiplied with my eigenvalues and then multiply this matrix with the eigenvectors

    \begin{bmatrix}17&16&16\\16&17&16\\16&16&17\end{bm  atrix} - \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}<br />
* \begin{bmatrix}x_{\lambda}\\x_{\lambda}\\x_{\lambd  a}\end{bmatrix}=0

    edit

    but the first and second matrix should be within brackets
    Follow Math Help Forum on Facebook and Google+

  8. #8
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Kind of. Solve this system:

    \begin{bmatrix}16&16&16\\16&16&16\\16&16&16\end{bm  atrix}\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatr  ix}=\begin{bmatrix}0\\0\\0\end{bmatrix}.

    This corresponds to \lambda=1. You get it down to

    \begin{bmatrix}1&1&1\\0&0&0\\0&0&0\end{bmatrix}\be  gin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}=\begi  n{bmatrix}0\\0\\0\end{bmatrix}.

    Set x_{2}=s, x_{3}=t. Then x_{1}=-s-t, and you have

    \begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}=s\  begin{bmatrix}-1\\1\\0\end{bmatrix}+t\begin{bmatrix}-1\\0\\1\end{bmatrix}.

    Therefore, your eigenvectors corresponding to \lambda=1 are what?
    Last edited by Ackbeet; November 5th 2010 at 05:52 AM. Reason: Misspelling of lambda.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    May 2010
    Posts
    16
    Lets see if I understand this correctly, to diagonolize my matrix (A) I need a diagonalizing matrix (P) and a Lambda*I matrix (D)

    A=PDP^{1}

    After finding my eigenvalues (1) I know that my D matrix is:
    \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}<br />

    I also know that AP = PD
    And that P has to be a nxn matrix

    After doing all the calculations I have gotten
    x=\begin{bmatrix}x{_1}\\x{_2}\\x{_3}\end{bmatrix}=  x_{2}\begin{bmatrix}-1\\x1\\x0\end{bmatrix}x_{3}\begin{bmatrix}-1\\0\\1\end{bmatrix}

    But from here i dont know how to move on, Are not x_{}2 and x_{3} my eigenvectors?

    And dont i need a third eigenvetctor?

    After all this, i know that my matrix has to be orthonormalized, can I orthonormalize my matrix after i have found it or should i do this with my eigenvectors?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,393
    Thanks
    1327
    In order that the matrix be diagonalizable, there must be 3 independent eigenvectors. If the only eigenvalue for this matrix is 1 and there are only two independent eigenvectors, then the matrix is NOT diagonalizable. In order that this matrix be orthogonalizable, there must be either another eigenvector corresponding to eigenvalue 1, independent of the two you have found, or there must be another eigenvalue.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Right. In the OP, there was another eigenvalue mentioned: 49/33. So you get two eigenvectors from the eigenvalue 1, and one eigenvector from the eigenvalue 49/33.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Response to # 9:

    Your matrix D will be as follows:

    D=\begin{bmatrix}1&0&0\\0&1&0\\0&0&49/33\end{bmatrix}.

    You construct your P from the orthonormalized eigenvectors. Let \mathbf{x}_{1} and \mathbf{y}_{1} be the eigenvectors corresponding to \lambda=1, and let \mathbf{z}_{49/33} be the eigenvector corresponding to \lambda=49/33. And let's assume that these are already-normalized eigenvectors. Then the matrix P (or P^{-1}, you'll have to double-check) is given by

    P=[\mathbf{x}_{1}\;\mathbf{y}_{1}\;\mathbf{z}_{49/33}].
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Newbie
    Joined
    May 2010
    Posts
    16

    eigenvalues

    So you mean that i dont have 1 as the only eigenvalue?
    Because 49/33 can be eliminated to 1 with gauss-elimination too.

    I am also a bit confused since you wrote in post #2 that 49/33 was not an eigenvalue.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Whoops. Maybe it would help if I read my own posts, sometimes! Well, you tell me. What are the eigenvalues? We have 1 as an eigenvalue, of algebraic and geometric multiplicity 2.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Newbie
    Joined
    May 2010
    Posts
    16
    Again I don't quite follow you since my english isn't so good, does of algebraic and geometric multiplicity 2 mean that I can multiply the 1 and use a multiple of it?
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Diagonalize unitary matrix
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: December 9th 2009, 08:08 PM
  2. Finding Eigenvectors to diagonalize
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: November 11th 2009, 02:17 AM
  3. How do I orthogonally diagonalize a matrix?
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: July 23rd 2009, 10:33 AM
  4. Orthogonally diagonalize the matrix
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: April 22nd 2009, 02:39 AM
  5. Orthogonally diagonalize the symmetric matrix
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: May 15th 2008, 07:26 AM

Search Tags


/mathhelpforum @mathhelpforum