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Math Help - Help with critical points

  1. #16
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    Re: Help with critical points

    Is it something to do with eigenvalues? Like Ax is an eigenvector of A or something? Or is it that Ax is equal to -(Ax \cdot x) ? If that's the case, wouldn't that make the original equation be  f(x) = (-(Ax \cdot x) \cdot x) e^{||x||^{2}}? Wouldn't that make it a scalar being on one side of the dot product and a vector on the other? This wouldn't work.
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  2. #17
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    Re: Help with critical points

    Never mind. It seems that  Ax = -(x^{T}Ax)x would work in the equation. But is that the right answer? I'm not sure.
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  3. #18
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    Re: Help with critical points

    Once I find what Ax is for the function to have a critical point, I need to calculate the Hessian in very simple terms as well. I tried some examples with numbers to check whether Ax = -(Ax\cdot x)x made the function have a critical point. but it doesn't seem to be the case.
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  4. #19
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    Re: Help with critical points

    So this is what I've come up with at this point:

    \nabla f(x) =0 = (A + (Ax \cdot x)I)x

    Knowing that for nonzero x, for that equation to be zero, it must hold that  det(A+(Ax \cdot x)I) = 0.

    This is true if the formula  det(A- \lambda I) = 0.

    Thus,  \lambda = -(Ax \cdot x).

    Then x there is a critical point of  f(x) \iff Ax = \lambda x where  \lambda is an eigenvalue of A.

    Is this accurate? It's the best I can come up with.

    I really need to finish this, as it's on my Final review for my Calc 3 class and I have my final tomorrow!

    Thanks for all your help so far!
    Last edited by tubetess123; December 8th 2011 at 10:33 PM.
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  5. #20
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    Re: Help with critical points

    change to an eigenvector basis of A which has real eigenvectors and real positive eigen values.

    Then (DX,X) = Li(Xi^2)......summation convention

    then f = Li(Xi^2)(e^(XiXi))

    ex: f = [L1(X1^2)+L2(X2^2)][e^(X1^2+X2^2)

    Now take d/dX1 = 0 and get X1 x positive terms =0 => X1=0
    same for d/dx2 = 0 and get X2=0

    The 0 vector is also the zero vector in original basis which is the critical point

    Note: If you don't switch to eigenvector basis you get

    (Ax,x) = A11(x1x1)+2A12(x1x2)+A22(x1x2) and good luck taking partials of f.

    Also, I assumed eigenvalues of a real symmetric matrix are positive. If not, its just more uninteresting, but doable, algebra.
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  6. #21
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    Re: Help with critical points

    So, let me make sure I understand what you mean:

    You're saying diagonalize the Symmetric matrix by using the equation  Q^{T}AQ where Q is an orthogonal matrix such that that expression gives a diagonal matrix?

    Then substitute that in for x so that I know that Ax \cdot x = \sum_i \lambda _{i} x_{i}^{2}?

    Is that what you mean?
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  7. #22
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    Re: Help with critical points

    Wait, I looked it up it I think you mean I should represent A as  A = QBQ^{T} where Q is the eigenvector basis of A, and B is a diagonal matrix with it's values being the eigenvectors of A.

    Is that what I need to do?
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  8. #23
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    Re: Help with critical points

    Quote Originally Posted by tubetess123 View Post
    Wait, I looked it up it I think you mean I should represent A as  A = QBQ^{T} where Q is the eigenvector basis of A, and B is a diagonal matrix with it's values being the eigenvectors of A.

    Is that what I need to do?
    Almost. The point is that if you switch to an eigenvector basis, Ax becomes Dx where D is a diagonal matrix of eigenvalues and the eq becomes Dxe^(llxll^2, I did make one mistake, eigenvalues are real but not positive. So result of setting partials to zero may be different. The point is the partials are now easy. Try it with a 2x2 matrrix.
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  9. #24
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    Re: Help with critical points

    So how do I switch it to an eigenvector basis? I feel like I should show how I did that. Or at least know how that's done. And why does Ax become Dx?
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  10. #25
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    Re: Help with critical points

    I don't know how to enter matrices with latex, but this is what I did. I'll enter them with wolfram alpha notation.

    A = {{1,2},{2,1}}

    Eigenvectors are: (-1,1) for eigenvalue -1. (1,1) for eigenvalue 3.

    So D would be {{-1,0},{0,3}}.

    But then what do I use the eigenvectors for? Are the pointless? Or do they have to do with how I make A into a diagonal matrix of eigenvalues?

    I guess I really just don't get how we arrived at D, as in what the formula is. And how we can just say that Ax = Dx.
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  11. #26
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    Re: Help with critical points

    Quote Originally Posted by tubetess123 View Post
    So how do I switch it to an eigenvector basis? I feel like I should show how I did that. Or at least know how that's done. And why does Ax become Dx?
    Assume you have switched to an eigen vector basis. The representation of A in that basis is D. Then in two dim eigenvector system using x,y for coordinates of the vector, the equation become)s:

    f = (ax^2 +by^2)e^(x^2+y^2) and take partials wrt x and y and set them to 0 and see what happens. In principle, having found x and y you can transform them back to components in original coordinate system, but I don't think it will come to that.

    The background principle here is, if I have a vector and Matrix for a given basis of V, what are they in a different basis? ie, if A and x are in one basis, and A' x' are in a different basis, then Ax=A'x' is the same vector expressed in different bases. You have to look up representations. I did't prove that llxll = llx'll, the length of a vector is the same in either coordinate system.
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  12. #27
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    Re: Help with critical points

    So when proving I don't need to show that I'm representing them in a different basis? And do I need to alter x at all to make it x'?
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  13. #28
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    Re: Help with critical points

    the vector x is the same in either basis. its representation by coordinates is different in different bases.
    for example: u = x1e1 + x2e2 =x1'e1' + x2'e2'

    Don't bother with primed coordinates. just make a note you are switching to eigenvector basis and the components are wrt that basis. The tip-off for switching to an eigenvector basis was that A is given as real and symmetric.

    First solve the problem in the simple intelligible form:

    f= [(ax^2+by^2]e^(x^2+y^2), x,y components of x in eigenvector system and a,b eigenvalues.

    to see what happens. Then you can clean it up. I assume you can takle partials wrt x and y and set them equal to 0?
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  14. #29
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    Re: Help with critical points

    Yep I think I've got it from here. I'm going to write it out and then if I have any more trouble I'll post back!

    Thank you so much for you help!
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  15. #30
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    Re: Help with critical points

    Ok I took all the partials, but couldn't it be true that one or all of the eigenvalues is zero for the partial to be zero?
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