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Post by Admin on Mar 30, 2017 11:28:00 GMT
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Gary
GaryVasco
Posts: 3,352 
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Post by Gary on Apr 9, 2017 22:55:42 GMT
Vasco My answer was considerably more tortured, but I think I reached a similar result. I had a look at your answer and it looks like I should study it and rewrite my own. nh.ch7.ex5.pdf (96.67 KB) (rewritten) In your answer in your calculation of the RHS second equation $J = ... = (1/2)((\xi , -\xi), (\eta, \eta))$, I wonder if the (1/2) should be $\frac{1}{\sqrt{2}}$ or $\frac{\sqrt{2}}{2}$. How do you reach the inferences in your last sentence? Gary |
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Post by Admin on Apr 11, 2017 6:42:10 GMT
Gary
You are right about the square root.
Also the characteristic equation should be
Lambda squared minus root (two)/2 times lambda times (xi+eta) plus (xi times eta)=0.
Vasco
PS I'm writing this on my phone and I can't find the backslash.
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Post by Admin on Apr 13, 2017 13:52:14 GMT
Gary
I have corrected my algebra in my solution and re-published. I have also looked at your solution. Although it is unnecessary to calculate the sum and product of the roots of the characteristic equation, since if we have a quadratic equation
$az^2+bz+c=0$,
the sum of its roots are -b/a and the product is c/a, if you do the calculation as you have done it does come out right without discarding the $1/\sqrt{2}$.
Also, on page 1 of your document in the second and third lines from the bottom, $\lambda_1$ and $\lambda_2$ should both be written as $\lambda$.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Apr 13, 2017 16:41:32 GMT
Vasco,
Thank you. I found the mistake in my algebra. I had never seen or had forgotten the simple formulas for the sum and product of the roots of a quadratic.
You wrote:
But I see that you have done the same thing (aside from my computing mistake) in calculating the characteristic equation from the Jacobian without considering the multiplier 1/2. What is the operative rule? That the Jacobian is a set of ratios that remains unchanged by division by a common factor? I will wait to repost until I understand this point.
Gary
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Post by Admin on Apr 13, 2017 19:08:10 GMT
Vasco, Thank you. I found the mistake in my algebra. I had never seen or had forgotten the simple formulas for the sum and product of the roots of a quadratic. You wrote: But I see that you have done the same thing (aside from my computing mistake) in calculating the characteristic equation from the Jacobian without considering the multiplier 1/2. What is the operative rule? That the Jacobian is a set of ratios that remains unchanged by division by a common factor? I will wait to repost until I understand this point. Gary Gary Sorry, that's my first version, I forgot to upload my corrected version - I've done it now. Vasco |
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Post by Admin on Apr 15, 2017 16:38:36 GMT
Vasco,
I have rewritten Ex. 5. I understand how you got your answer, but I'm interested to know what you think this means regarding the implications of rotation of a linear transformation for the calculation of eigenvalues.
Gary
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Post by Admin on Apr 15, 2017 16:42:24 GMT
Gary
In your document you say: "...we could have treated M as the Jacobean."
This is only true as far as calculating the determinant of J is concerned. As you say rotation does not affect the area, which is represented by the determinant of the Jacobean.
The characteristic equations for J and M are different and therefore so are the eigenvalues
The eigenvalues are equal to ξ and η only if there is no net rotation.
The answers to your questions in the final paragraph of your document are therefore both no, for the reasons given above.
Vasco
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Post by Admin on Apr 15, 2017 16:49:39 GMT
Vasco,
Sorry for the mess in the previous. I didn't realize I was in the message editor where I can't edit. This might be better.
Thank you. I'm trying to grasp the lesson that Needham must have had in mind. We find in this case of rotation by $\pi/4$ that
$J=R^{\pi/4}_a.M$, where M=((ξ,0),(0,η))
det(M)=det[J]=λ1λ2=ξη
and
ξ and η are generally not the same as the eigenvalues of the rotated matrix $J$,
so
det(J−λI)≠det(M−λI)
We can find the product of the roots of the characteristic equation (eigenvalues) by either (a) solving the characteristic equation for the λ's or (b) Viete's method that λ1λ2=c/a.
Do you think the point, besides providing some practice with the Jacobian and linear transformations, is just to acquaint us with this special case? Or is it to show that rotation will generally (but not in this case) distort the shape of an image curve? It does not appear that the purpose was to show that the determinant is constant under rotation.
Gary
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Post by Admin on Apr 16, 2017 7:25:25 GMT
Gary
I'm not sure what the point of exercise 5 is other than to show that the topological multiplicity at a point $a$ does not depend on the rotation. Eigenvectors and Eigenvalues are only mentioned on pages 158-160 and 170-171. I'll give it some more thought.
Vasco
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