Results 1 to 6 of 6

Math Help - Finding the eigenvalues of a 4x4 matrix?

  1. #1
    Newbie
    Joined
    Nov 2008
    Posts
    3

    Finding the eigenvalues of a 4x4 matrix?

    Hi!

    I need help finding the eigenvalues of a 4x4 matrix. Been working with the same problem for 2 days now. I just can't solve it. Help me pls!
    The matrix (A) is:

    <br /> <br />
\left( \begin{array}{cccc} 1 & -3 & 2 & -1 \\ -3 & 9 & -6 & 3 \\ 2 & -6 & 4 & -2 \\ -1 & 3 & -2 & 1\end{array} \right)<br /> <br />


    (Notice that the matrix is symmetric, if that is to any help)

    I have come this far (If I'm on the right track).

    det <br /> <br />
\left( \begin{array}{cccc} 1-\lambda & -3 & 2 & -1 \\ -3 &<br />
9-\lambda & -6 & 3 \\ 2 & -6 & 4-\lambda & -2 \\ -1 & 3 & -2 & 1-\lambda \end{array} \right)<br /> <br />
= 0


    How do I solve for L from this? I would appreciate any help.

    Thanks!
    /Gordina
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Your matrix should be:

    \begin{bmatrix}{\lambda}-1&3&-2&1\\3&{\lambda}-9&6&-3\\-2&6&{\lambda}-4&2\\1&-3&2&{\lambda}-1\end{bmatrix}

    This gives a characteristic polynomial of:

    {\lambda}^{4}-15{\lambda}^{3}={\lambda}^{3}({\lambda}-15)=0

    Now, solve this quartic and you have your eigenvalues.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Hi,

    Another approach :
    Quote Originally Posted by Gordina View Post
    I need help finding the eigenvalues of a 4x4 matrix. Been working with the same problem for 2 days now. I just can't solve it. Help me pls!
    The matrix (A) is:
    <br /> <br />
\left( \begin{array}{cccc} 1 & -3 & 2 & -1 \\ -3 & 9 & -6 & 3 \\ 2 & -6 & 4 & -2 \\ -1 & 3 & -2 & 1\end{array} \right)<br /> <br />

    (Notice that the matrix is symmetric, if that is to any help)
    Let v=\begin{pmatrix}1\\-3\\2\\-1 \end{pmatrix}. Note that A=\begin{pmatrix}v&-3v&2v&-v \end{pmatrix} so \mathrm{Im}(A)=\mathrm{Span}(\{v\}) which implies \dim\left(\mathrm{Im}(A)\right)=\dim\left(\mathrm{  Span}(\{v\})\right)=1. As \dim\left(\mathrm{Im}(A)\right)+\dim\left(\ker(A)\  right)=\dim(\mathbb{R}^4)=4, one has \dim\left(\ker(A)\right)=3>0 hence 0 is an eigenvalue of A.

    The spectral theorem tells us that because A is symmetric, it is diagonalizable hence :

    (1) _ the multiplicity of an eigenvalue of A equals the dimension of the corresponding eigenspace ;
    (2) _ the sum of the dimension(s) of the eigenspace(s) of A equals \dim(\mathbb{R}^4)=4.

    Thanks to (1), we get that the multiplicity of the eigenvalue 0 is \dim\left(\ker(A-0\cdot\mathrm{I}_3)\right)=\dim\left(\ker(A)\right  )=3. As \dim\left(\ker(A)\right)=3\neq4, (2) implies that one eigenvalue \alpha (of multiplicity 1) remains to be found. This can easily be done using the equality \mathrm{tr}(A)=\sum_{\lambda\in\mathrm{Sp(A)}}\lam  bda where \mathrm{Sp}(A)=\{0,\alpha\} is the spectrum of A and \mathrm{tr}(A) is the trace of A.

    If this proof is correct it could probably be generalized to any matrix A=\begin{pmatrix} v_1 v& v_2v&\cdots& v_n v\end{pmatrix} where v=\begin{pmatrix} v_1 \\ v_2\\ \vdots \\v_n\end{pmatrix}\in\mathbb{R}^n. It would show that the two eigenvalues of A are 0 (multiplicity = n-1) and \mathrm{tr}(A) (multiplicity = 1).
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Nov 2008
    Posts
    3

    Looks promising!

    To galactus:
    Are you sure lambda should be to the left in the matrix? My book says different. I guess you'r right, since you surely know better than me

    I also wan't to know how you got the characteristic polynomial of the matrix. Did you use cofactor expansion? Or is there an easier way? (Please say there's an easier way.)


    Anyway, the two answers upove seems intressting, since both characteristic polynomials and diagonalization is a part of my course. I will look at this tomorrow, and get back to you. Now I need some sleep...

    Thanks a lot for the help!


    Edit:
    galactus, I now noticed that you also have the rows in different orders than me. What am I doing wrong here?
    Last edited by Gordina; November 6th 2008 at 03:02 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Nov 2008
    Posts
    3

    Finding eigenvalues of a 4x4 matrix.

    I asked this question in the homework forum, but desided to move it here, since I'm not in a hurry.

    My problem:
    How do I find the eigenvalues of this matrix:

    <br />
\begin{bmatrix}1&-3&2&-1\\-3&9&-6&3\\2&-6&4&-2\\-1&3&-2&1\end{bmatrix}<br />

    (Notice that it is symmetric, if that helps)

    As I understand I need to solve
     {\lambda} from:

    det <br />
\begin{bmatrix}1-{\lambda}&-3&2&-1\\-3&9-{\lambda}&-6&3\\2&-6&4-{\lambda}&-2\\-1&3&-2&1-{\lambda}\end{bmatrix}<br />
= 0

    The only way I know how to do this is by cofactor expansion. But that would take all day for me. There's got to be a faster way.


    galactus said:
    " Your matrix should be:

    \begin{bmatrix}{\lambda}-1&3&-2&1\\3&{\lambda}-9&6&-3\\-2&6&{\lambda}-4&2\\1&-3&2&{\lambda}-1\end{bmatrix}


    This gives a characteristic polynomial of:

    {\lambda}^{4}-15{\lambda}^{3}={\lambda}^{3}({\lambda}-15)=0 "


    Why is galactus using a different matrix, and how did he get it? And how did he get the characteristic polynomial of that matrix?

    I've been working on this problem for several days now. I just can't solve it. I would be so happy if anyone could help me.


    /Gordina


    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,569
    Thanks
    1409
    Quote Originally Posted by Gordina View Post
    I asked this question in the homework forum, but desided to move it here, since I'm not in a hurry.

    My problem:
    How do I find the eigenvalues of this matrix:

    <br />
\begin{bmatrix}1&-3&2&-1\\-3&9&-6&3\\2&-6&4&-2\\-1&3&-2&1\end{bmatrix}<br />

    (Notice that it is symmetric, if that helps)

    As I understand I need to solve
     {\lambda} from:

    det <br />
\begin{bmatrix}1-{\lambda}&-3&2&-1\\-3&9-{\lambda}&-6&3\\2&-6&4-{\lambda}&-2\\-1&3&-2&1-{\lambda}\end{bmatrix}<br />
= 0

    The only way I know how to do this is by cofactor expansion. But that would take all day for me. There's got to be a faster way.


    galactus said:
    " Your matrix should be:

    \begin{bmatrix}{\lambda}-1&3&-2&1\\3&{\lambda}-9&6&-3\\-2&6&{\lambda}-4&2\\1&-3&2&{\lambda}-1\end{bmatrix}


    This gives a characteristic polynomial of:

    {\lambda}^{4}-15{\lambda}^{3}={\lambda}^{3}({\lambda}-15)=0 "


    Why is galactus using a different matrix, and how did he get it? And how did he get the characteristic polynomial of that matrix?

    I've been working on this problem for several days now. I just can't solve it. I would be so happy if anyone could help me.


    /Gordina


    You are using det(A- \lambda I)= 0 which is certainly what I would tend to do do. Galactus is using det(\lambda I- A)= 0, just multiplying the entire characteristice equation by -1. Of course, those two equations have the same solutions. Perhaps Galactus just doesn't like writing " -\lambda"! I suspect he got the characteristic equation by doing exactly what you said would "take all day"- expanding the determinant.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding a value in a matrix with distinct eigenvalues
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 28th 2009, 01:16 PM
  2. Finding eigenvalues and eigenvectors of a real matrix
    Posted in the Advanced Algebra Forum
    Replies: 7
    Last Post: August 18th 2009, 06:05 AM
  3. Finding the eigenvalues of a 3x3 matrix
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: July 30th 2009, 02:32 PM
  4. Eigenvalues, eigenvectors, finding an orthogonal matrix
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: May 5th 2009, 11:06 AM
  5. Replies: 1
    Last Post: May 6th 2008, 07:24 AM

Search Tags


/mathhelpforum @mathhelpforum