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Lin Alg - Eigenvector Existence proof

  1. Apr 25, 2006 #1
    Lin Alg - Eigenvector Existence proof (More of a proof ?, than eigenvector ?)

    Here is the question from my book:

    Show that If [itex]\theta \in \mathbb{R}[/itex], then the matrix

    [tex]A = \left(\begin{array}{cc}cos \theta & sin \theta \\ sin \theta & -cos \theta \end{array}\right)[/tex]

    always has an eigenvector in [itex]\mathbb{R}^2[/itex], and in fact that there exists a vector [itex]v_{1}[/itex] such that [itex]Av_{1} = v_{1}[/itex]

    Now I am a little caught up on one thing here:

    Is this saying that for any value of [itex]\theta \in \mathbb{R}[/itex], A has an eigenvector in [itex]\mathbb{R}^2[/itex] ?? (imo it does)

    Because I am working a simple proof by contradiction, and if I can just choose a value of [itex]\theta[/itex] I would be done. (Except for the second part, but imo, that is simple to do, as I would just need to show that there is some eigenvalue = 1.)

    Last edited: Apr 25, 2006
  2. jcsd
  3. Apr 25, 2006 #2
    I guess I should be more specific in what I am doing.

    I am letting theta be in R, then I am assuming that A has no eigenvectors. Thus for all [itex] X \in \mathbb{R}^2 AX \neq \lambda X [/itex] for any [itex]\lambda[/itex] Then I am letting X = t(0 1) that t is the transpose.
    Then I get [itex]AX = t(sin \theta \ \ cos\theta) [/itex]
    and [itex]\lambda X = t( 0 \ \ \lambda )[/itex]
    Now if [itex]\theta[/itex] takes on certain values, then [itex]AX = \lambda X[/itex] which is a contradiction (that is, if theta can be any value in R)
    Last edited: Apr 25, 2006
  4. Apr 25, 2006 #3


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    Your approach is wrong. You must show the existence of eigenvectors for all values of theta, not just certain values of theta.

    Since you have to solve the eigenvalue equation, why not just go a step further an find the eigenvectors (it's not long at all) ? If you get stuck along the way, someone will bail you out.
  5. Apr 25, 2006 #4


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    You're supposed to prove that for ALL [itex]\theta[/itex] in [itex]\mathbb{R}[/itex], the corresponding matrix A has an eigenvector; you've only shown that only for SOME [itex]\theta \in \mathbb{R}[/itex], the corresponding matrix A has an eigenvector.

    Do you know the geometric interpretation of these matrices A? If so, then the whole thing would be easy, and you'd know what eigenvector to look for. Try investigating. Look at the eight points on the unit circle that are at some multiple of 45 degrees. Now consider the matrices you get when you take [itex]\theta[/itex] to be some multiple of 45 degrees.

    Or here's a way to almost always find an eigenvector: Pick a vector (x 1)T. Apply the matrix A to it, and you'll get a resulting vector:

    [tex]\left (\begin{array}{c}{f(x,\theta )\\g(x,\theta )}\end{array}\right )[/tex]

    For some functions f and g. Remember you're asked to find, given the angle [itex]\theta[/itex] an eigenvector, so you want to find x such that [itex]f(x, \theta ) = x[/itex], [itex]g(x, \theta ) = 1[/itex] since you already expect the eigenvalue to be 1. This procedure will only fail when [itex]\theta[/itex] is an integer multiple of 360 degrees.
  6. Apr 25, 2006 #5
    I had first approached it that way, but I could just not seem to get anywhere, now I got it. Thanks!
  7. Apr 25, 2006 #6


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    Now that you've figured out the answer for yourself, you should know that this matrix gives a reflection in the line through the origin which makes an angle [itex]\theta / 2[/itex] with the positive x-axis. Any vector in this line is obviously an eigenvector with eigenvalue 1.
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