1. Let V be any inner product space. Let u, v be any two vectors in V.
Prove the following: ||u+v||^2 + ||u-v||^2 = 2*||u||^2 + 2||v||^2.

2. By rotating the coordinate axes in the positive direction,
eliminate the cross product xy term. Then identify the new equation.

x^2 + xy + y^2 = 1.

Thanks so much for your help! I'm asking these problems to assist me
with understanding all the material; it's not a substitute for hw.
Please respond with a complete answer asap (preferably before 8pm).
Thanks again.

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Clarification of Question by alison28-ga on 26 Apr 2005 16:22 PDT

please respond to this by midnight... I offered a greater amt of money
for the fast reply. Thanks again for your time.

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Clarification of Question by alison28-ga on 26 Apr 2005 18:23 PDT

Just wanted to remind you that this is *not* for anything that's
graded. ...im just trying to prepare for a final in this class, and
I'm having trouble w/ these ones.

Also, if you can't do one of them, just give me what you can do. Thank you!

Hi!!


1. Let V be any inner product space. Let u, v be any two vectors in V.
Prove the following: ||u+v||^2 + ||u-v||^2 = 2*||u||^2 + 2||v||^2.

Recall the properties of the inner product:
1. <x;y> = <y;x>
2. <x;a*y> = a*<x;y> and <a*x;y >= a*<x;y>
3. <x;(y+z)> = <x;y> + <x;z> and <(x+y);z> = <x;z> + <y;z>
4. <x;x> >= 0 with equality iif x = 0

Recall also the definition of norm:
||x|| = sqrt(<x;x>) or ||x||^2 = <x;x>

Now we can start:
||u+v||^2 = <(u+v);(u+v)> = 
          = <(u+v);u> + <(u+v);v> =
          = <u;u> + <u;v> + <u;v> + <v;v> =
          = ||u||^2 + 2*<u;v> + ||v||^2

In the same way we find that:
||u-v||^2 = <(u-v);(u-v)> = <(u-v);(u+(-1)*v)> =
          = <(u-v);u> + <(u-v);(-1)*v> = <(u-v);u> - <(u-v);v> =
          = <u;u> - <u;v> - (<u;v> - <v;v>) =
          = <u;u> - <u;v> - <u;v> + <v;v> =
          = ||u||^2 - 2*<u;v> + ||v||^2

Joining both equalities we have that:
||u+v||^2 + ||u-v||^2 = ||u||^2 + 2*<u;v> + ||v||^2 + ||u||^2 -
                        - 2*<u;v> + ||v||^2 =
                      = 2*||u||^2 + 2*||v||^2


I hope that this helps you. I am posting this because you need at
least partial answers asap but I am continuing working on the second
problem. If you find something unclear please post a request for an
answer clarification and I will gladly give you further assistance on
this.

Regards.
livioflores-ga

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Request for Answer Clarification by alison28-ga on 27 Apr 2005 00:18 PDT

Could you try to post the answer to 2) by 5:00am? thanks so much for
helping me to understand these (there will be similar ones on the
final).

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Clarification of Answer by livioflores-ga on 27 Apr 2005 02:08 PDT

Hi again!!

2. By rotating the coordinate axes in the positive direction,
eliminate the cross product xy term. Then identify the new equation.

x^2 + xy + y^2 = 1


we have an equation of the form Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 with:
A=1
B=1
C=1
D=E=0
F=-1

To identify the conic we must evaluate the discriminant B^2 - 4AC :
B^2 - 4AC = 1 - 4*1*1 = 1-4 = -3 < 0 , then we have an Ellipse, a
Circle, a Point, or Nothing.

Since B is different to Zero a rotation is required. We have that A=C
then the rotation angle is a=PI/4 (45º) (otherwise tan(2a) = (B-A)/C
):

Perform the rotation:
x = u cos(a) - v sin(a) = u/sqrt(2) - v/sqrt(2) = (u-v)/sqrt(2)
y = u sin(a) + v cos(a) = u/sqrt(2) + v/sqrt(2) = (u+v)/sqrt(2)


Replace in the main equation:

x^2+xy+y^2-1 = 1/2*(u^2-2uv+v^2)+ 1/2*(u^2-v^2) + 1/2*(u^2+2uv+v^2)-1 =
             = 1/2*[u^2-2uv+v^2 + u^2-v^2 + u^2+2uv+v^2] - 1 =
             = 1/2*[3u^2+v^2] - 1 =
             = 3/2*u^2 + 1/2*v^2 -1 = 0

Then the new equation is:
3/2*u^2 + 1/2*v^2 = 1

or

u^2/U^2 + v^2/V^2 = 1 with U = sqrt(2/3) and V = sqrt(2)

It is clearly an elipse.         


For references on this topic see (from more simple to more complete
explanations and examples):
"7.1 Alternative Characterization":
http://www.geom.uiuc.edu/docs/reference/CRC-formulas/node27.html

"SparkNotes: Conic Sections: Axis Rotation":
http://www.sparknotes.com/math/precalc/conicsections/section5.rhtml

"Ellipse Algebra - 3DSoftware.com":
See the 'Unrotating an Ellipse' section at the bottom of the page.
http://www.3dsoftware.com/Math/PlaneCurves/EllipseAlgebra/

"Translations and Rotations":
http://www.pen.k12.va.us/Div/Winchester/jhhs/math/lessons/calculus/rotate.html

"Rotation of Conics into Standard Form":
http://math.etsu.edu/MultiCalc/Chap3/Chap3-1/part4.htm

"7.2 The General Quadratic Equation":
http://www.geom.uiuc.edu/docs/reference/CRC-formulas/node28.html#SECTION01720000000000000000

"Rotation of Axes":
http://www.sci.ccny.cuny.edu/math/Courses/math20200/rotatedconics.pdf


I hope that this helps you. Feel free to request for a clarification
if you need it.

Regards.
livioflores-ga

Thanks, that's what I got as well. I just figured the problem out
right before you posted, but it's good to have reassurance.