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Find the Eigenvectors and eigenvalues of this matrix

  1. Aug 4, 2010 #1
    I'm trying to find the Eigenvectors and eigenvalues of this matrix:

    Code (Text):
    [
    0 0 0 0
    0 0 0 0
    0 0 0 1
    0 0 1 0
    ]
    I get 0, 1, and -1 as my eigenvalues.

    Starting with 0, I solve for reduced row echelon form and get the matrix:

    Code (Text):
    [
    0 0 1 0 0
    0 0 0 1 0
    0 0 0 0 0
    0 0 0 0 0
    ]
    My question is, and maybe my brain is just fried, but I'm confused at to what my vector should look like.

    I know x3=0 and x4=0, but what about x1,2?

    My guess: I would just assign them an arbitrary variable r?

    For example x=<r,r,0,0>, let r=1, <1,1,0,0>?

    Another question: My characteristic equation is L^2(L^2-1)=0; Are my roots 0, 0, -1, 1 or 0, -1, 1?
     
    Last edited: Aug 4, 2010
  2. jcsd
  3. Aug 4, 2010 #2

    lanedance

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    Homework Helper

    Re: Eigenvectors

    you're right x3 = x4 = 0, x1 & x2 are arbitrary, so why not look at
    (1,0,0,0)
    (0,1,0,0)
    any vector with x3 = x4 = 0 can be written as a linear combination of those 2
     
  4. Aug 4, 2010 #3
    Re: Eigenvectors

    You seem to be on the right track, but just a few details are a little off.

    You solved your characteristic equation, and (correctly) got:

    [tex] \lambda^2(\lambda^2 - 1) = 0 [/tex].

    Now the eigenvalues are the roots of this equation, with multiplicity equal to their multiplicity as roots. Here zero is a repeated root (twice). What that means for us is that 0 will be an eigenvalue with multiplicity 2; that is, there will be two eigenvectors associated with the eigenvalue 0.

    You saw this when you solved for the eigenvalues. You wound up with a solution to the homogenous system where x1 and x2 were both associated with rows of zeros. That's because there are two eigenvectors associated with the eigenvalue 0. You don't use the same variable "r" for both x1 and x2; they don't have to be equal to satisfy the solution.

    Instead you have one vector [0,1,0,0] and another [1,0,0,0] which are eigenvectors.

    Strictly speaking, your eigenvalues are "0 (with multiplicity 2), -1, and 1" but if it helps you to keep it straight to think of them as "0,0,-1,1", I don't see any harm in it.
     
  5. Aug 5, 2010 #4

    lanedance

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    Homework Helper

    Re: Eigenvectors

    just to add there is a difference between multiplicities
    - algrebraic (power of term in charactersitic equation)
    - geometric (number of eigenvectors for that eigenvalue)

    and geomtric is always less than or equal to algebraic
     
  6. Aug 5, 2010 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Re: Eigenvectors

    The matrix is
    [tex]\begin{bmatrix}0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 9 & 1\\ 0 & 0 & 1 & 0\end{bmatrix}[/tex]

    which, as you say, has 1, -1, and 0 as eigenvalues.

    hgfalling's statement: "Here zero is a repeated root (twice). What that means for us is that 0 will be an eigenvalue with multiplicity 2; that is, there will be two eigenvectors associated with the eigenvalue 0." isn't quite correct. As lanedance says, that means that the algebraic multiplicity is 2. The geometric multiplicity, the number of independent eigenvectors corresponding to eigenvalue 0, may be 1 or 2.

    Here's the way to determine the eigenvectors directly from the definition:
    If <w, x, y, z> is an eigenvector with eigenvalue 0, then
    [tex]\begin{bmatrix}0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\end{bmatrix}\begin{bmatrix}w \\ x \\ y \\ z\end{bmatrix}= \begin{bmatrix}0 & 0 & z & y\end{bmatrix}= \begin{bmatrix}0\\ 0 \\ 0 \\ 0\end{bmatrix}[/tex]

    which is the same as the four equations 0= 0, 0= 0, z= 0, and y= 0. We must have y and z 0 but w and x can be anything: any eigenvector with eigenvalue 0 is of the form <w, x, 0, 0>= <w, 0, 0, 0>+ <0, x, 0, 0>= w<1, 0, 0, 0>+ x<0, 1, 0, 0>
    [tex]\begin{bmatrix}w \\ x \\ 0 \\ 0\end{bmatrix}= \begin{bmatrix}w \\ 0 \\ 0 \\ 0 \end{bmatrix}+ \begin{bmatrix}0 \\ x \\ 0 \\ 0 \end{bmatrix}= w\begin{bmatrix}1 \\ 0 \\ 0 \\ 0\end{bmatrix}+ x\begin{bmatrix}0 \\ 1 \\ 0 \\ 0\end{bmatrix}[/tex].

    For eigenvalue 1, we must have
    [tex]\begin{bmatrix}0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\end{bmatrix}\begin{bmatrix}w \\ x \\ y \\ z\end{bmatrix}= \begin{bmatrix}w\\ x \\ y \\ z\end{bmatrix}[/tex]
    That gives four dependent equations which give a one-parameter solutions.

    similarly for eigenvalue -1.
     
  7. Aug 5, 2010 #6
    Re: Eigenvectors

    Right. Listen to them.
     
  8. Aug 6, 2010 #7

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Re: Eigenvectors

    Although, in this particular case, the algebraic multiplicity happens to be equal to the geometric multiplicity!

    Other examples are:
    [tex]\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}[/tex]
    which has 1 as a triple eigenvalue and has <1, 0, 0> , <0, 1, 0>, and <0, 0, 1> as independent eigenvectors. Algebraic multiplicity= geometric multiplicity= 3.

    [tex]\begin{bmatrix}1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}[/tex]
    which again has 1 as a triple eigenvalue but only has <1, 0, 0> and <0, 0, 1> as independent eigenvectors. Algebraic multiplicity= 3, geometric multiplicity= 2.

    Note that [tex]\begin{bmatrix}1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}0 \\ 1\\ 0\end{bmatrix}= \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix} [/tex]
    is not an eigenvector.

    Finally,
    [tex]\begin{bmatrix}1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}[/tex]
    has 1 as a triple eigenvector but its eigenvectors are all multiples of <1, 0, 0>. Algebraic multiplicity= 3, geometric multiplicity= 1.
     
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