Results 1 to 7 of 7

Math Help - polynomial, eigenvalues and eigenvectors

  1. #1
    Banned
    Joined
    Jan 2009
    Posts
    58

    polynomial, eigenvalues and eigenvectors

    How do I find the characteristic polynomial, eigenvalues, and eigenvectors of the following matrix:
     <br />
\begin{pmatrix}2 & -2 & 3\\0 & 3 & -2\\0 & -1 & 2\end{pmatrix}<br />
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2008
    From
    France
    Posts
    1,458
    Quote Originally Posted by jennifer1004 View Post
    How do I find the characteristic polynomial, eigenvalues, and eigenvectors of the following matrix:
     <br />
\begin{pmatrix}2 & -2 & 3\\0 & 3 & -2\\0 & -1 & 2\end{pmatrix}<br />
    Hi

    t is an eigenvalue of matrix A iff there exists a non-zero vector X such that AX = tX or (A-tI)X = 0 where I is the identity matrix (1 on the diagonal ; 0 everywhere else)
    This means that the determinant of A-tI must be equal to 0

    Set the determinant of  <br />
\begin{pmatrix}2-t & -2 & 3\\0 & 3-t & -2\\0 & -1 & 2-t\end{pmatrix}<br />
to 0 and solve for t to find the eigenvalues
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member rubix's Avatar
    Joined
    Apr 2009
    Posts
    49
    Quote Originally Posted by jennifer1004 View Post
    How do I find the characteristic polynomial, eigenvalues, and eigenvectors of the following matrix:
    \begin{pmatrix}2 & -2 & 3\\0 & 3 & -2\\0 & -1 & 2\end{pmatrix}<br />
    Hi there jennifer1004

    for easiness let us denote given matrices as A.

    A = \begin{pmatrix}2 & -2 & 3\\0 & 3 & -2\\0 & -1 & 2\end{pmatrix}<br />

    first to find characteristic polynomial:

    det ( A - xID)

    1) subtract x from diagonal (the one going from left to right) of the matrix A; leave rest as it is.

    \begin{pmatrix}2-x & -2 & 3\\0 & 3-x & -2\\0 & -1 & 2-x\end{pmatrix}<br />

    2) take the determinant of matrices obtained from # 1

    det A = [ (2-x) (3-x) (2-x) ] - [ (-1) (-2) (2-x) ]

    3) set the result you get from # 2 equal to zero. This is your characteristic function.

    secondly, to find e-values:

    4) solve for x in your characteristic function. These value(s) of x is your e-value(s).

    lastly, to find e-vectors:

    (A - xID) V = 0 ; where ID is the identify matrix and V is the e-vector that you are looking for.

    5) If you obtained more than one e-value then you'll have to do follow following step more than once - one e-value at a time. Say, you had two e-values namely p and q. Go over p and then over q. I'll do it for p.

    6) go back to #1 and plug p in there. You'll get

    \begin{pmatrix}2-p & -2 & 3\\0 & 3-p & -2\\0 & -1 & 2-p\end{pmatrix}<br />

    {remember p is a number...it is a e-value}

    7) find A - xID

    \begin{pmatrix}2 & -2 & 3\\0 & 3 & -2\\0 & -1 & 2\end{pmatrix} - p \begin{pmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{pmatrix} =

    8) use the formula (A - xID) V = 0 and solve

    (A - xID)  \begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}


    hope that helps!
    Last edited by rubix; April 17th 2009 at 02:56 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member Twig's Avatar
    Joined
    Mar 2008
    From
    Gothenburg
    Posts
    396

    hi

    Easiest way to solve the determinant is probably using a cofactor expansion across column 1.

     (2-\lambda)((3-\lambda)(2-\lambda) -2)
    Now solve the quadratic equation and you have your eigenvalues.

    And as rubix already explained, you find your eigenvectors by finding a basis for the nullspace of the matrix  A-\lambda_{1}I , where  \lambda_{j} \, j=1,2,3 are the eigenvalues.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Banned
    Joined
    Jan 2009
    Posts
    58
    Thank you for your explanations. I will read them over more carefully very shortly. I am trying to focus and absorb so many different types of info right now that it is so hard for me to just understand everything. It is a little overwhelming, but thank you for all the time you spent helping me.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Banned
    Joined
    Jan 2009
    Posts
    58
    I got the polynomial of x^{3}-7x^{2}+14x-8

    Eigenvalues of 2, 4, 1

    Eigenvectors <-1,1,1> for 1, <1,0,0> for 2 and <7/2, -2, 1> for 4. I'm not sure if these are even correct but would these be my final eigenvectors, or am I supposed to multiply them by the corresponding eigenvalue. For example, <1, 0, 0> for 2 would be <2, 0, 0> instead as my final eigenvector?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Banned
    Joined
    Jan 2009
    Posts
    58
    I think I got this now. I believe I have to multiply the eigenvectors by the corresponding eigenvalue. My final eigenvectors should be:

    for 2 <2 0 0>
    for 4 <14 -8 4>
    for 1 <-1 1 1>

    ?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. eigenvectors/eigenvalues
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: April 20th 2011, 11:38 AM
  2. Eigenvalues and Eigenvectors
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 9th 2011, 04:57 AM
  3. polynomial, eigenvalues, eigenvectors, diagonal matrix
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 20th 2009, 09:19 PM
  4. Eigenvalues/Eigenvectors
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 7th 2009, 10:15 PM
  5. Eigenvalues/Eigenvectors
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: December 11th 2007, 07:13 PM

Search Tags


/mathhelpforum @mathhelpforum