Results 1 to 5 of 5

Math Help - eigenvalues/eigenvectors of a 3x3 matrix

  1. #1
    Newbie
    Joined
    May 2010
    Posts
    2

    eigenvalues/eigenvectors of a 3x3 matrix

    i have a 3x3 matrix

    <br />
\begin{pmatrix}-2 & -8 & -12\\1 & 4 & 4\\0 & 0 & 1\end{pmatrix}<br />

    i got the eigenvalues of 2, 1, and 0.
    im having a big problem with how to get the corresponding eigenvectors

    if anyone can help me that would be great!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member AllanCuz's Avatar
    Joined
    Apr 2010
    From
    Canada
    Posts
    384
    Thanks
    4
    Quote Originally Posted by eigenvectorisnotfun View Post
    i have a 3x3 matrix

    <br />
\begin{pmatrix}-2 & -8 & -12\\1 & 4 & 4\\0 & 0 & 1\end{pmatrix}<br />

    i got the eigenvalues of 2, 1, and 0.
    im having a big problem with how to get the corresponding eigenvectors

    if anyone can help me that would be great!
    For every eigenvalue there is a corresponding eigenvector. So we need to find our eigenvectors 3 times!

    How we do this is place lambda on the diagonal in the matrix! In other words,

     \begin{pmatrix} {-2 - \lambda }& -8 & -12\\ 1  & { 4- \lambda } & 4\\0 & 0 & {1 - \lambda } \end{pmatrix}

    Let's pick  \lambda = 0

     \begin{pmatrix} {-2}& -8 & -12\\ 1 & { 4 } & 4\\0 & 0 & {1} \end{pmatrix}

    Now we need to row reduce!

     12 R_3 + R_1 --> R_1

     \begin{pmatrix} {-2}& -8 & 0\\ 1 & { 4 } & 4\\0 & 0 & {1} \end{pmatrix}

     R_2 - 4R_3 --> R_2

     \begin{pmatrix} {-2}& -8 & 0\\ 1 & { 4 } & 0\\0 & 0 & {1} \end{pmatrix}

     2 R_2 + R_1 --> R_1

     \begin{pmatrix} {0}& 0 & 0\\ 1 & { 4 } & 0\\0 & 0 & {1} \end{pmatrix}

    At this point I don't think we can reduce any more.

    Now we know there exists a set of vectors such that

     \begin{pmatrix} {0}& 0 & 0\\ 1 & { 4 } & 0\\0 & 0 & {1} \end{pmatrix} \begin{pmatrix} V_1 \\ V_2 \\V_3 \end{pmatrix} = 0

    Carry out the multiplication and find your values. There might be trivial solutions but that's okay!

    Find more at: Eigenvalues and Eigenvectors
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,792
    Thanks
    1532
    Or, use the definition of "eigenvalue" and "eigenvector".

    If you are right that 2 is an eigenvalue, then there must exist a vector, \begin{pmatrix}x \\ y \\ z\end{pmatrix}, x, y, z not all 0, such that
    \begin{pmatrix}-2 & -8 & -12 \\ 1 & 4 & 4 \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix} = \begin{pmatrix}-2x- 8y- 12z \\ x+ 4y+ 4z \\ z\end{pmatrix} = \begin{pmatrix}2x \\ 2y \\ 2z\end{pmatrix}.

    That gives the three equations -2x-8 y- 12z= 2x, x+ 4y+ 4z= 2y, and z= 2z.
    Of course, that last equations tells us that z= 0 so the first two become -2x- 8y= 2x, or 8y= -4x so that x= -2y, and x+ 4y= 2y and, again, x= -2y. The fact that there exist an infinite number of solutions to those equations confirms that 2 is an eigenvalue. Any eigenvector corresponding to eigenvalue 2 is of the form \begin{pmatrix}-2y \\ y \\ 0\end{pmatrix}= y\begin{pmatrix}-2 \\ 1 \\ 0\end{pmatrix}.

    Now, to find the eigenvectors corresponding eigenvalues 1 and 0, just solve -2x- 8y- 12z= x, x+ 4y= y, z= z, and -2x- 8y- 12z= 0, x+ 4y= 0, z= 0.

    On the contrary, eigenvectors are great fun! (Not to mention extremely useful.)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member AllanCuz's Avatar
    Joined
    Apr 2010
    From
    Canada
    Posts
    384
    Thanks
    4
    Quote Originally Posted by HallsofIvy View Post
    Or, use the definition of "eigenvalue" and "eigenvector".

    If you are right that 2 is an eigenvalue, then there must exist a vector, \begin{pmatrix}x \\ y \\ z\end{pmatrix}, x, y, z not all 0, such that
    \begin{pmatrix}-2 & -8 & -12 \\ 1 & 4 & 4 \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix} = \begin{pmatrix}-2x- 8y- 12z \\ x+ 4y+ 4z \\ z\end{pmatrix} = \begin{pmatrix}2x \\ 2y \\ 2z\end{pmatrix}.

    That gives the three equations -2x-8 y- 12z= 2x, x+ 4y+ 4z= 2y, and z= 2z.
    Of course, that last equations tells us that z= 0 so the first two become -2x- 8y= 2x, or 8y= -4x so that x= -2y, and x+ 4y= 2y and, again, x= -2y. The fact that there exist an infinite number of solutions to those equations confirms that 2 is an eigenvalue. Any eigenvector corresponding to eigenvalue 2 is of the form \begin{pmatrix}-2y \\ y \\ 0\end{pmatrix}= y\begin{pmatrix}-2 \\ 1 \\ 0\end{pmatrix}.

    Now, to find the eigenvectors corresponding eigenvalues 1 and 0, just solve -2x- 8y- 12z= x, x+ 4y= y, z= z, and -2x- 8y- 12z= 0, x+ 4y= 0, z= 0.

    On the contrary, eigenvectors are great fun! (Not to mention extremely useful.)
    Is this typically a faster approach then the row reductions I have above?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    May 2010
    Posts
    2
    now i finally got it!
    what i did wrong was I did not really get that (A-I lambda) = 0
    thanks to you both, you both were really of great help
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: December 6th 2010, 01:00 PM
  2. Replies: 4
    Last Post: April 25th 2010, 03:49 AM
  3. [SOLVED] Eigenvalues and eigenvectors of a 3x3 matrix A? A unknown?
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: March 17th 2010, 08:58 AM
  4. Eigenvectors and eigenvalues of a matrix
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: May 6th 2008, 08:29 AM
  5. Replies: 1
    Last Post: May 6th 2008, 07:24 AM

Search Tags


/mathhelpforum @mathhelpforum