Never mind. It seems that would work in the equation. But is that the right answer? I'm not sure.
Is it something to do with eigenvalues? Like Ax is an eigenvector of A or something? Or is it that Ax is equal to ? If that's the case, wouldn't that make the original equation be ? 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.
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 made the function have a critical point. but it doesn't seem to be the case.
So this is what I've come up with at this point:
Knowing that for nonzero x, for that equation to be zero, it must hold that .
This is true if the formula .
Thus, .
Then x there is a critical point of where 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!
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.
So, let me make sure I understand what you mean:
You're saying diagonalize the Symmetric matrix by using the equation 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 ?
Is that what you mean?
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.
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.
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.
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?