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Use Lagrange multipliers to find the eigenvalues and eigenvectors of a matrix

  1. Nov 21, 2012 #1
    1. The problem statement, all variables and given/known data

    Use Lagrange multipliers to find the eigenvalues and eigenvectors of the matrix

    [itex]A=\begin{bmatrix}2 & 4\\4 & 8\end{bmatrix}[/itex]

    2. Relevant equations

    ...

    3. The attempt at a solution

    The book deals with this as an exercise. From what I understand, it says to consider the function [itex]f(x,y) = \frac{1}{2}(A[x,y]) \cdot [x,y][/itex], with the assumption that [itex]A[/itex] is symmetric (which is the case here).

    It then asks what the gradient of the function is, which is [itex]\nabla f(x,y) = A[x,y][/itex].

    It then asks to restrict [itex]f[/itex] to the region [itex]S=\{[x,y] \in \mathbb{R}^2 : x^2 + y^2 \leq 1\}[/itex]. Then, it then states that there must exists a vector [itex][x,y] \in S[/itex] and a real number [itex]\lambda \neq 0[/itex] such that [itex]A[x,y] = \lambda [x,y][/itex], and claiming that finding the maxima and minima of [itex]f[/itex] constrained to [itex]S[/itex] will give the eigenvalues and eigenvectors of [itex]A[/itex].

    Why is this true? To be more precise, why do the eigenvalues and eigenvectors only exists in the unit disk? And, how do we know there exists such a [itex]\lambda \neq 0[/itex] that [itex]A[x,y] = \lambda [x,y][/itex]? I tried Lagrange multipliers, but I could only verify that [itex]A[x,y] = 2 \lambda [x,y][/itex].

    Thanks.
     
  2. jcsd
  3. Nov 21, 2012 #2

    Dick

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    The reason this works is that since the matrix is symmetric you know the two eigenvectors are orthogonal. Call them e1 and e2 with eigenvectors λ1 and λ2. And you actually want to extremize on the unit circle x^2+y^2=1 (not the unit disk). A vector on the unit circle is given by v=a*e1+b*e2 where a^2+b^2=1, yes? Now you should be able to compute that A(v).v=λ1*a^2+λ2*b^2. With me so far? Now if λ1>λ2 then clearly the max of A(v).v is where a=1, b=0 and the min is where a=0, b=1. So the extrema of the Lagrange problem will give you the eigenvectors e1 and e2. To solve the lagrange problem try and eliminate x from the equations and look at the equation for y and then eliminate y since it can't be zero and give you an eigenvector. Find possible values of lambda. And sure, there are eigenvectors outside of the unit circle, but they are all multiples of the ones on the unit circle.
     
    Last edited: Nov 21, 2012
  4. Nov 21, 2012 #3

    Dick

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    BTW, this is a bit harder than finding the eigenvalues and eigenvectors the usual way using linear algebra. You might want to try that easier way first so you know what to expect. That's what I did.
     
    Last edited: Nov 21, 2012
  5. Nov 25, 2012 #4
    thanks for the reply.

    I still find the concept odd, but it is a bit more clear, thanks. I think I just need time for this to settle in my mind, maybe.

    A few questions, though.

    One, when we are referring to the eigenvectors as [itex]e_1[/itex] and [itex]e_2[/itex], are we referring to them as [itex][1,0][/itex] and [itex][0,1][/itex], the standard basis vectors of [itex]\mathbb{R}^2[/itex]? If not, then my understanding of equality of [itex](Av) \cdot v = a^2 \lambda_1 + b^2 \lambda_2[/itex] goes out the window, as my work for that relied on that.

    Since the rest of my questions rely on the eigenvectors being such standard basis vectors, I'll wait for a response.

    thanks

    yeah, I agree, haha
     
  6. Nov 25, 2012 #5

    Dick

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    No, e1 and e2 aren't the standard basis. But you can select them to be orthonormal eigenvectors of A. They are orthogonal because A is symmetric and you can normalize them so |e1|=|e2|=1. So e1.e1=e2.e2=1 and e1.e2=0.
     
    Last edited: Nov 25, 2012
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