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Homework Help: Linear Algebra

  1. Sep 16, 2007 #1
    1. The problem statement, all variables and given/known data

    The problem is given in the attachment below. The relevant information is provided under the "Stereo Audio" heading. Additionally, I have a couple of pieces of extra information not listed in the Stereo Audio section:

    M is the component of (column vector) [R, L] on (column vector) [1, 1]
    (This notation means that R and L are two entries in a column vector. The "R" being the upper entry and the "L" being the lower entry. The same applies for the 1 and 1)

    D is the component of (column vector) [R, L] on (column vector) [-1, 1]

    For the first problem, it asks me to express (M^2) + (D^2) in terms of L and R. Am I correct in writing (M^2) + (D^2) = [(L+R)^2]/4 +[(L-R)^2]/4?

    Attached Files:

  2. jcsd
  3. Sep 16, 2007 #2
    Is there a reason that my attachment has not been approved? How long does it usually take?
  4. Sep 16, 2007 #3
    Last edited by a moderator: May 3, 2017
  5. Sep 17, 2007 #4


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    It takes a while for any link to another site to be approved. (Some brave soul with really really good virus protection has to check it out.)

    I don't see that there is really any "Linear Algebra" involved here.

    Yes, if M= (L+ R)/2 and D= (L-R)/2 then M2+ D2= (L+R)2/4+ (L-R)2/4. It would be a good idea, I think, to multiply those squares out and combine the fractions.
  6. Sep 17, 2007 #5
    Yeah, thats what it seemed to me as well.

    Additionally, I can't seem to be able to picture exactly what this scenario would look like graphed. First, we are given M= (L+ R)/2, which appears to be half the length of the vector [R, L]. However, at the same time I am told M is the component of vector [R, L] on [1, 1] (basically, a projection of [R, L] onto the [1, 1] vector) However, it seems these cannot be true at the same time, at least in my mind. So is it possible to graph these based on the given information?
    Last edited: Sep 17, 2007
  7. Sep 17, 2007 #6


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    No, (L+ R)/2 is certainly not "half the length of the vector [R,L]"! That would be [itex](1/2)\sqrt{L^2+ R^2}[/itex].
    Drawing a right triangle diagram, you should be able to see that the projection of vector u on vector v has length [itex]|u|sin\theta [/itex] where [itex]\theta[/itex] is the angle between u and v. [itex]Since u.v= |u||v|sin\theta[/itex], we can write that as u.v/|v|. In the particular case that u= [L, R] and v= [1, 1] that is [itex](L+ V)/\sqrt{2}[/itex].
  8. Sep 17, 2007 #7
    So is M= (L+ R)/2 a sum of two vectors or the sum of their magnitudes?
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