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Stuck with eigenvectors.

  1. Apr 6, 2005 #1
    The eigenvalues of the matrix [tex] \left(\begin{array}{cc}0 & \frac{1}{2}\\ \frac{1}{2} & 0\end{array}\right) [/tex] are [itex] \lambda_1 = \frac{1}{2} [/itex] and [itex] \lambda_2 = -\frac{1}{2} [/itex]
    The problem here is that I have no idea of how to calculate the eigenvectors. Could some one please explain me, in detail, how do I find the eigenvectors?
    Thanks in advance. (sorry for posting the thread twice)
    Last edited: Apr 6, 2005
  2. jcsd
  3. Apr 6, 2005 #2
    Those aren't the eigenvalues...

    You should get [itex]\lambda_1 = 0, \; \lambda_2 = 1/2[/itex].

    Edit: Okay, now that you've fixed your post, solve the equations [itex]Av = \lambda_1 v[/itex] and [itex]Av = \lambda_2 v[/itex].
    Last edited: Apr 6, 2005
  4. Apr 6, 2005 #3
    There is no zero eigenvalue, as the transformation of this matrix is 1-1 on R2. Specifically, the transformation includes an inversion of one axis, as it maps a scaling of the positive x-axis to the positive y-axis and vice-versa for the positive y-axis (Remember that the columns of the matrix form a basis for the image of the transformation). Ignoring the scaling factor of 1/2, it is then easy to see that two distinct eigenvectors are (1,1) (rotate the positive x-axis into the positive y-axis, then invert the old y-axis) and (1,-1) by the same visualization is mapped to its negative. Noting that the scaling factor is 1/2, we then find the eigenvalues are 1/2 and -1/2 as you found.
    Last edited: Apr 6, 2005
  5. Apr 6, 2005 #4
    In the case that the transformation of the matrix is not very nice, you will want an algebraic way of getting eigenvectors. This can be done by simply plugging each eigenvalue you get back into the equation [tex](A - I\lambda)v = 0[/tex] where A is the matrix, lambda is the eigenvalue, I is the identity matrix, and v is an eigenvector for lambda and A, and solve for v. For example, when [tex]\lambda = \frac{1}{2}[/tex], you will want to solve the equation:
    [tex]\left(\begin{array}{cc}-\frac{1}{2} & \frac{1}{2}\\ \frac{1}{2} & -\frac{1}{2}\end{array}\right) \left(\begin{array}{c}v^1\\ v^2\end{array}\right) = 0[/tex]
    Note that you are solving for a one-dimensional subspace (not just a single vector), so you will have one free variable.
    Last edited: Apr 6, 2005
  6. Apr 6, 2005 #5
    This is where I get stuck. If I solve the equation [tex]\left(\begin{array}{cc}-\frac{1}{2} & \frac{1}{2}\\ \frac{1}{2} & -\frac{1}{2}\end{array}\right) \left(\begin{array}{c}v^1\\ v^2\end{array}\right) = 0[/tex], I get the next system of equations:
    [tex] -1/2v^1 + 1/2v^2 = 0 ... (1) [/tex]
    [tex] 1/2v^1 - 1/2v^2 = 0 ... (2) [/tex]
    and I think the solution for this system is [tex] v^1 = v^2 [/tex], so I donĀ“t know how to interpret this.
  7. Apr 6, 2005 #6
    correct. It means that your eigenvector is of the form

    [tex](v_1 \ , v_1)^T = v_1(1 \ , \ 1)^T[/tex]

    ie. a basis for the eigenvectors corresponding to the eigenvalue [itex] 1 /2 [/itex] is

    [tex](1\ ,\ 1)^T.[/tex]
  8. Apr 6, 2005 #7
    So, this means that the first eigenvector will be (1/2, 1/2)?
  9. Apr 6, 2005 #8
    that is fine as well. Using [itex](1 \ , \ 1), \ ( 1/2\ ,\ 1/2), [/itex] or [itex](1001412\ , \ 1001412)[/itex] are all equivalent.
  10. Apr 6, 2005 #9
    Ok, i got that. Just help me with this other example. The eigenvalues of the matrix [tex] \left(\begin{array}{cc}5 & -3\\ -3 & 5\end{array}\right) [/tex] are [itex] \lambda_1 = 2 [/itex] and [itex] \lambda_1 = 8 [/itex], and, according to my book, the eigenvectors are [itex] (\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}) [/itex] and [itex] (-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}) [/itex]. I dont get how did they got them.
  11. Apr 6, 2005 #10
    Well, what did you get when you tried to find them? What did you do to find your answers?
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