1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding the eigenvectors of a matrix A

  1. Nov 27, 2015 #1
    1. The problem statement, all variables and given/known data
    [tex]
    A = \begin{bmatrix}
    2 & 1 & 0\\
    0& -2 & 1\\
    0 & 0 & 1
    \end{bmatrix}
    [/tex]

    2. Relevant equations


    3. The attempt at a solution

    The spectrum of A is [itex] \sigma (A) = { \lambda _1, \lambda _2, \lambda _3 } = {2, -2, 1 } [/itex]

    I was able to calculate vectors [itex]v_1[/itex] and [itex]v_3[/itex] correctly out of the vectors on the following page:
    http://www.wolframalpha.com/input/?i=eigenvectors+of+{{2,1,0},{0,-2,1},{0,0,1}}

    However, [itex]v_2[/itex] is giving me a headache. Using [itex] \lambda _1[/itex] to solve [itex]A - \lambda _1 I_3[/itex] gives me the matrix

    [tex]
    \begin{bmatrix}
    \stackrel{a}{0} & \stackrel{b}{1} & \stackrel{c}{0}\\
    0 & -4 & 1\\
    0 & 0 & -1
    \end{bmatrix}

    \stackrel{rref}{=}

    \begin{bmatrix}
    \stackrel{a}{0} & \stackrel{b}{1} & \stackrel{c}{0}\\
    0 & 0 & 1\\
    0 & 0 & 0
    \end{bmatrix}
    [/tex]

    In my head this would produce a zero eigenvector since

    [tex]\begin{cases} b = 0 \\ c = 0 \\ (a = ?) \end{cases}[/tex]

    This is of course nonsense. I'm probably interpreting the row-reduced matrix wrong, but what is it exactly that I'm not understanding? Does it have something to do with the fact that on every row a = 0?
     
  2. jcsd
  3. Nov 27, 2015 #2
    To maybe answer my own question, (someone correct me if I'm wrong):

    If each row in the above matrix represents an equation, I could theoretically make [itex]a = s, s \in \mathbb{R} [/itex], since we're solving for the null space, meaning that since [itex]a[/itex] is always zero, it is equal to each the components of a zero vector, and we could add aything to either side of the equation and it would still hold, transforming the system of equations into:

    [tex]
    \begin{cases}
    a = s\\
    b=0\\
    c=0
    \end{cases}
    [/tex]

    Is this the solution?
     
  4. Nov 27, 2015 #3
    a can then be any value you could choose 1 to normalize.
     
  5. Nov 27, 2015 #4
    Got it. Thank you for confirming my suspicions.
     
  6. Nov 30, 2015 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    l would prefer to use the basic definitions of "eigenvalues" and "eigenvectors". Saying that "2" is an eigenvalue means that there exist a non-zero vector (in fact an entire subspace of them), (x, y, z), such that
    [tex]\begin{bmatrix} 2 & 1 & 0 \\ 0 & -2 & 1 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}x \\y \\ z \end{bmatrix}= \begin{bmatrix}2x+ y \\ -2y+ z \\ z \end{bmatrix}= 2\begin{bmatrix}x \\ y \\ z \end{bmatrix}[/tex]
    which gives the three equations 2x+ y= 2x, -2y+ z= 2y, z= 2z. The first equation is the same as y= 0 and the last z= 0. But there is no condition on x. Any vector of the form (x, 0, 0)= x(1, 0, 0) is an eigenvector corresponding to eigenvalue 2.

    Similarly, the eigenvectors corresponding to eigenvalue -2 must satisfy 2x+ y= -2x, -2y+ z= -2y, z= -2z. We get z= 0 from the last equation but see that, then, any y satisfies the second and the first says y= -4x. So (x, -4x, 0)= x(1, -4, 0) is an eigenvector corresponding to eigenvalue -2.

    Finally, eigenvectors corresponding to eigenvalue 1 must satisfy 2x+ y= x, -2y+ z= y, z= z. The last equation is always true, the second says z= 3y, and the first y= -x so that z= 3y= -3x. An eigenvector corresponding to eigenvalue 1 is (x, -x, -3x)= x(1, -1, -3).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Finding the eigenvectors of a matrix A
  1. Finding Eigenvectors (Replies: 3)

  2. Finding eigenvectors. (Replies: 4)

Loading...