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 eigenvectors of [[1,-1,-1],[-1,1,-1],[-1,-1,1]]

  1. Jul 15, 2012 #1
    he eigenvalues of the 3x3 matrix [[1,-1,-1],[-1,1,-1],[-1,-1,1]] are 2,2, and -1.
    how can i compute the eigenvectors?
    for the case lambda=2, for example, i end up with an augmented matrix [[-1,-1,-1,0],[-1,-1,-1,0],[-1,-1,-1,0]] so i'm stuck at this point.


    much appreciated.
     
  2. jcsd
  3. Jul 15, 2012 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    So, you need to solve the linear system
    [tex] x_1 + x_2 + x_3 = 0\\
    x_1 + x_2 + x_3 = 0\\
    x_1 + x_2 + x_3 = 0 [/tex]
    There are lots of solutions. In fact, you should be able to find two linearly independent solution vectors, corresponding to the double eigenvalue 2.

    RGV
     
  4. Jul 15, 2012 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I prefer to work from the basic definitions (perhaps I just never learned these more sophisticated methods!):
    Saying that 2 is an eigenvalue of this matrix means there exist a non-zero vector such that
    [tex]\begin{bmatrix}1 & -1 & -1 \\ -1 & 1 & -1 \\ -1 & -1 & 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}x - y- z\\ -x+ y- z \\ -x- y+ z\end{bmatrix}= \begin{bmatrix}2x \\ 2y\\ 2z\end{bmatrix}[/tex]
    which gives the three equations x- y- z= 2x, -x+ y- z= 2y, -x- y+ z= 2z which are, of course, equivalent to -x- y- z= 0, -x- y- z= 0, -x- y- z= 0. Those three equations are the same. We can, for example, say that z= -x- y so that any vector of the form <x, y, -x- y>= <x, 0, -x>+ <0, y, -y>= x<1, 0, -1>+ y<0, 1, -1> is an eigenvector. Notice that the eigenvalue, 2, not only has algebraic multiplicity 2 (it is a double root of the characteristic equation) but has geometric multiplicity 2 (the space of all corresponding eigenvalues is 2 dimensional).

    Similarly, the fact that -1 is an eigenvalue means there are x, y, z, satisfying x- y- z= -x, -x+ y- z= -y, -x- y+ z= -z which are, of course, equivalent to 2x- y- z= 0, -x+ 2y- z= 0, -x- y+ 2z= 0. If we subtract the second equation from the first, we eliminate z to get 3x- 3y= 0 so y= x. Putting that into the third equation, 2x+ 2z= 0 so z= -x.
    Any eigenvector corresponding to eigenvalue -1 is of the form <x, x, -x>= x<1, 1, -1>.
     
  5. Jul 15, 2012 #4
    by that reasoning, can i not have <1,-1,0> and <-1,1,0> as my two solutions for eigenvalue 2? but wolframalpha says i need to have the case where y=0.

    also, i think the solution for eigenvalue -1 is <1,1,1>
     
  6. Jul 16, 2012 #5

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Your two listed vectors (for eigenvalue 2) are just multiples of each other. You need two *linearly independent* eigenvectors, such as <1,-1,0> and <1,0,-1>, or <0,-1,1> and <1,-1/2, -1/2>, etc. There are infinitely many possible pairs of vectors <x1,y1,z1> and <x2,y2,z2> that are linearly independent and satisfy the equation x+y+z=0. Any such pair will do.

    RGV
     
    Last edited: Jul 16, 2012
  7. Jul 16, 2012 #6
    yes, i had thought of that, but found it uncomfortable that of all the many possibilities, both my professor and wolframalpha chose the cases y=0 and z=0 as solutions, so i was wondering what made these two special compares to the others.
     
  8. Jul 17, 2012 #7

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    There is nothing special about these choices, except for the fact that they both have one component = 0 so are, in a sense, the simplest possible. However, you could equally take x=0 and y=0 or x=0 and z=0.

    RGV
     
  9. Jul 17, 2012 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    If you have a vector that depends upon parameters, say, <x, y, -x- y> as I have above, then choosing x= 0, y= 1 gives you <0, 1, -1> and choosing x= 1, y= 0 gives <1, 0, -1>. That is, in effect, the same as writing <x, y, -x- y>= <x, 0, -x>+ <0, y, -y>= x<1, 0, -1>+ y<0, 1, -1>, showing that any such vector is a linear combination of <1, 0, -1> and <0, 1, -1>.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Finding eigenvectors of [[1,-1,-1],[-1,1,-1],[-1,-1,1]]
  1. Finding eigenvectors (Replies: 13)

  2. Finding eigenvectors? (Replies: 1)

Loading...