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Inner Product

  1. Oct 15, 2005 #1
    Oh, its algebra time again!

    A Question reads:
    let ||u|| =1 ||v|| = 2 ||w|| = 3^0.5 (or root 3) <u,v>=-1
    <u,w>=0 and <v,w>=3

    Given this information, who that u + v = w

    I gave it my best show.

    i know that ||u|| (im going to write |u| for slimpicity) ... i know that |u| = <u,u> = 1 and |v| = <v,v> = 2 and that |w| = <w,w> = root 3.

    I did some arithmatic with the given data, but i just cannot seem to isolate u and v.

    What I am now trying to do is work from converting an inner products to the vectors, however, I dont even think that is possible. Infact, I looked through two different text books and I still couldn't find it.

    Could somebody give me a head start, for where to being. Dont give me too much because I want to work this out on my own.

    Thanks
     
  2. jcsd
  3. Oct 15, 2005 #2

    Gokul43201

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    Consider the vector u+v-w. Being a linear combination of these three vectors, in must lie in any subspace that contains these vectors. Now what can you infer, if you find that the inner product of this vector with each of u, v, and w turns out to be 0 ?
     
  4. Oct 15, 2005 #3

    AKG

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    This is wrong. Given an inner product, the norm is usually defined:

    [tex]\|v\| \equiv \sqrt{\langle v,\, v\rangle }[/tex]
     
  5. Oct 15, 2005 #4

    Gokul43201

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    Also, in terms of vectors in Euclidean space, the inner product <u,v> = |u| |v| cos(A), where A is the angle between u and v. In the above case, u and w are orthogonal vectors whose lengths are given. The direction and magnitude of v can be easily found from the given data and the above definition. Knowing all three vectors (in polar representation), you can show that the required expression holds.
     
  6. Oct 15, 2005 #5
    Ahh Gokul43201! I showed it by the second method you told me. Thats geometry! but it works!

    so...

    1. <u w> = |u| |w| cos(A) A = 90deg
    2. <v w> = |v| |w| cos(A) A = 30deg
    3. <u v> = |u| |v| cos(A) A = 120deg

    So the angle between u and v is 120 deg.
    And so, u+v = w.. and the angle between w and v is 30deg, and the angle between w and u is 90deg, and this adds to 120deg.

    Therefore, u+v =w!

    and AKG... yes, that was a very silly mistake that I made. I don't know how i missed that.

    Furthermore, Gokul43201... on your first post, you were speaking about linear combinations.. " Now what can you infer, if you find that the inner product of this vector with each of u, v, and w turns out to be 0 ? " You mean that it is perpendicular right!
     
  7. Oct 15, 2005 #6

    AKG

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    I don't understand how your argument works playboy. Gokul, does what he wrote make sense? Anyways, you know that <x,x> = 0 if and only if x = 0. And you want to prove u+v = w, i.e. u+v-w = 0.
     
  8. Oct 16, 2005 #7

    Gokul43201

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    No playboy : just showing that the angles add up correctly means nothing...they have to do that. If you take any 3 vectors u,v,w with A being the angle between u and v and B being the angle between v and w, then the angle between u and w can only be one of A+B or A-B. This is true for any 3 vectors !

    The best way to prove it (ignore my first post, that is quite non-optimal - but no, the inference is not that it is perpendicular*) is as AKG says above. There's only one vector which has norm=0.

    If you have to go the geometrical way, it's a little more elaborate. If you've done Newtonian Mechanics and vector addition, you'll easily recall the formulas for the magnitude and direction of a resultant vector - which is nothing but the sum of the two vectors. You can find the resultant of u,v and show that it is w.

    * If <a,b> = 0, then a is orthogonal to b or one of the two vectors is a zero-vector. If some vector x satisfies <x,a> = <x,b> = <x,c> = 0 (with a,b,c non-zero) then either x must be orthogonal to the minimal space containing a,b,c or x=0. But the first is impossible since x lies in the minimal space containing a,b,c.
     
    Last edited: Oct 16, 2005
  9. Oct 16, 2005 #8
    oh thats right!

    Its to hard to show my geometry on here, but when you draw out the vectors, it does make sense.

    so i just solved for the direction, now i need the magnitude.

    i used the angel that I found between v and w, 30deg and the angel outside u and v which is 60deg... and used that sin law...

    sin(u) = sin(W)
    u w

    sin(30deg) = sin(60deg) and solve for w
    1 w

    and i get root 3!

    So, I found all the angels and showed the directions of the vectors, and now i just showed the magnitude! So this has to be right!
     
  10. Oct 16, 2005 #9

    AKG

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    Are you sure? What if the vectors are not coplanar?
     
  11. Oct 16, 2005 #10
    Yes, perhaps. But I still think that showing the magnitude of u+v using trig proves the question!

    I have another question that comes from the text book...

    b) If < , > and < , >' are two inner products on V that have equal associated norm functions, show that <u,v>=<u,v>' for all u and v.

    This is how i understand the question:

    < , > and < , >' are two DIFFERENT inner products in V.
    they have equal associated norm funtions: ||< , >|| = ||< , >'||
    Does that sound right?

    i mean, commen sense would tell me that <a,b> = <a,b>'

    can somebody give me a little push.

    Thanks.
     
  12. Oct 16, 2005 #11

    Gokul43201

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    Oops...of course !
     
  13. Oct 16, 2005 #12

    Gokul43201

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    No, how can that be ? The inner product is a csalar. What's the norm of a scalar ?

    What that line says is that given the two inner products <u,v> and <u,v>', you have ||u|| = sqrt(<u,u>) = sqrt(<u,u>') = ||u||' for all u.
     
  14. Oct 16, 2005 #13
    Isn't that still commen sense though? that if they have equal norm funtions, then obviously <u,v> and <u,v>' are going to be equal?
    I mean, further the question gives
    ||u|| = ||u||'
    ||v|| = ||v||'
    So wouldn't one conculde that <u,v> = <u,v>'
     
  15. Oct 17, 2005 #14

    Gokul43201

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    In math, you do not believe that something is true even if it makes sense that it should. Nothing is taken as a truth unless it can be proved from the set of axioms.

    Try finding the two norms of a vector like u+v and set them equal to each other...
     
    Last edited: Oct 17, 2005
  16. Oct 17, 2005 #15
    ........That is almost the same thing the prof told us on the first week of university...... and the last word we ever want to hear is "prove" or "show" :frown:
     
    Last edited by a moderator: Oct 17, 2005
  17. Oct 17, 2005 #16
    In the inner product section of the text book....

    the norm (length) is.... ||u|| = root <u,u>
    the distance is.... d(u,v) = ||u-v||

    if you have u+v, then the norm would be.... root u^2 + v^2 right...

    So what do you mean by finding TWO norms of a vector like u+v?
     
  18. Oct 17, 2005 #17

    Gokul43201

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    Look into the properties of inner products, specifically <a, b+c> = <a,b> + <a,c>

    Exploit this property with the suggestion I made. Calculate <u+v,u+v> and <u+v,u+v>' and compare terms.
     
  19. Oct 17, 2005 #18

    HallsofIvy

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    Are you under the impression that (u+v).(u+v)= u^2+ v^2??? I would hope that a beginning algebra student would know that that is not true for u and v numbers, much less a person who is taking linear algebra! What exactly do you mean by "u^2"? the dot product of u with itself? if so then
    the norm would be the root of (u^2+ 2u.v+ v^2).
     
  20. Oct 17, 2005 #19

    HallsofIvy

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    The fact that the norms are the same tells you that <a,a>= <a,a>' for any a in the vector space. It does not immediately tell you that <a,b>= <a,b>' for any a,b. You might try looking at <a+b,a+b>= <a+b,a+b>' and expanding.
     
  21. Oct 17, 2005 #20
    Hallsofivy... I should have been more clear on that! Next time I will!

    I was actually trying to refer to the distance in 3 space between two points u and v which is the root of (x-x0)^2 + (y-y0)^2 + (z-z0)^2 which is completely irrelevent to this question!

    Im just getting so confused with working with inner products right now. So im just going to spend some time working with simple questions with the properties right now.
     
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