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How to find eigenvectors of a matrix

  1. Dec 3, 2009 #1
    My textbook doesn't seem to explain it clearly enough for me to comprehend. But from what I can see, after getting the eigenvalues, you sub them back in to the lambdas that are in the matrix:

    [tex](\lambda I - A)x = 0[/tex]

    From here, you can solve for the system of equations with Gaussian elimination. After you solved it, the eigenvector is determined by the values of each unknown, where they take the coefficient of the parameters (if present).

    (quick example) So if I get a matrix and reduce it to RRE that says:

    [ 1 0 2 ]
    [ 0 1 3 ]
    [ 0 0 0 ]

    The the values are:

    X3 = t
    X2 = 3t
    X1 = 2t

    So the vectorspace I get from this is:
    [ 2 ]
    [ 3 ]
    [ 1 ]

    Or something?

    This is probably way wrong. Can someone please help explain to me how to calculate eigenvectors? I need it to solve one of my problems.


    ** I will personally send Christmas cards to anyone who helps me with understanding this (provided that you pm me your address)**
     
    Last edited: Dec 3, 2009
  2. jcsd
  3. Dec 3, 2009 #2
    Better yet, I'll post the example:

    [tex]
    A = \left[ \begin{array}{cc} -6 & 0.25 \\ 7 & -3 \end{array} \right] = \left[ \begin{array}{cc} \lambda+6 & -0.25 \\ -7 & \lambda+3 \end{array} \right]

    [/tex]

    [tex]
    \lambda^2+9\lambda+16.25
    [/tex]

    [tex]
    \lambda=-2.5
    [/tex]

    [tex]
    \lambda=-6.5
    [/tex]

    From here is where I need [itex]
    (\lambda I - A)x = 0
    [/itex], which yields:

    [tex]

    \left[ \begin{array}{cc} -2.5+6 & -0.25 \\ -7 & -2.5+3 \end{array} \right]
    [/tex]

    Now, I have tried using Gaussian elimination, but I get really screwy decimal numbers which I really don't think is right. Can someone take me through the process of calculating it and explain the steps and reasoning? I've checked online eigenvector calculators and it said the eigen vectors are:

    (1, -2) for -6.25
    (1, 14) for -2.25

    Not sure how to get to those results (or if they're even correct).

    Thanks.
     
  4. Dec 4, 2009 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I have always tended to go back to the definition. As iamsmooth says, [tex]A= \begin{bmatrix}-6 & .25 \\ 7 & -3\end{bmatrix}[/tex] has eigenvalues -2.5 and -6.5.

    Now, the definition of "eigenvalue" is that it is a number, [itex]\lambda[/itex] such that [itex]Av= \lambda v[/itex] is satisfied by some non-zero vector v. (And, since v= 0 is obviously a solution, it there exists non-zero solutions, there must exist an infinite number of them- the set of all eigenvectors corresponding to a given eigenvector forms a subspace.)

    That is, any eigenvector corresponding to eigenvalue -2.5 must satisfy
    [tex]\begin{bmatrix}-6 & .25 \\ 7 & -3\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= -2.5\begin{bmatrix}x \\ y\end{bmatrix}[/tex]

    [tex]\begin{bmatrix}-6x+ .25y \\ 7x- 3y\end{bmatrix}= \begin{bmatrix}-2.5x \\ -2.5y\end{bmatrix}[/itex]
    That gives the two equations -6x+ .25y= -2.5x and 7x- 3y= -2.5y. Adding 6x to both sides of first equation gives .25y= 3.5x or y= 14x. Adding 3y to both sides of the second equation gives 7x= .5y or, again, y= 14x.

    Those two equations are dependent and we cannot solve for specific values of x and y because, as I said, there exist an infinite number of eigenvectors for each eigenvalue. We can choose x to be whatever we want and solve for y, getting an eigenvector. For example, if we take x= 1, then y= 14 so <1, 14>, as iamsmooth got, is an eigenvector corresponding to eigenvalue -2.5.
     
    Last edited: Dec 4, 2009
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