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Math Help - Orthogonal sets

  1. #1
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    Question Orthogonal sets

    Is it possible to find a orthogonal set that is not linearly independent?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Orthogonal sets

    Quote Originally Posted by icanc View Post
    Is it possible to find a orthogonal set that is not linearly independent?
    An orthogonal subset S of an euclidean space is not linearly independent if and only if 0\in S .
    Last edited by FernandoRevilla; November 26th 2011 at 10:43 PM.
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  3. #3
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    Re: Orthogonal sets

    More generally, if v_1, v_2, v_3, ..., v_n are orthogonal vectors in an inner product space (we need the inner product to be able to define "orthogonal"), let a_1v_+ a_2v_2+ a_3v_3+ \cdot\cdot\cdot+ a_nv_n= 0 be a linear combination of the vectors that is equal to the 0 vector. Take the inner product of both sides of that with each vector in turn. For example, the inner product of each side with [itex]v_1[/tex] would be v_1\cdot(a_1v_1+ a_2v_2+ a_3v_3+ \cdot\cdot\cdot+ a_nv_n)= a_1v_1\cdot v_1+ a_2v_1\cdot v_2+ a_3\v_1\cdot v_3+ \cdot\cdot\cdot+ a_nv_1\cdot v_n v_1\cdot 0. The right side is obviously 0 while on the left side, every product except [itex]v_1\cdot\v_1[/itex] is 0. You did not say "orthonormal" so we cannot assume v_1\cdot v_1= 1 but it is some non-zero number (as FernandoRevill says, assuming the set does NOT contain the 0 vector), p, so we have a_1p= 0 and must conclude that a_1= 0.
    Last edited by HallsofIvy; November 27th 2011 at 02:13 PM.
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    Re: Orthogonal sets

    I am still a little confused. So, let's say that v_1= [3, 1, 1], v_2= [-1, 2, 1], and v_3 = [-1/2, -2, 7/2]. They are all orthogonal, I checked. Could you give me a similar example that would make the set not linearly independent?
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  5. #5
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    Re: Orthogonal sets

    sure, just add the 0-vector [0,0,0].

    the point being, if we have an orthogonal set in R^3 that is all non-zero vectors, say v1,v2,v3. then consider any linear combination:

    u = av1 + bv2 + cv3. if this is 0, then <v1,u> = 0, since <v1,0> = 0 (here <x,y> is just the dot product in R^3).

    but <v1,u> = <v1,av1> + <v1,bv2> + <v1,cv3> = a<v1,v1> + b<v1,v2> + c<v1,v3> = a<v1,v1> (by orthogonality).

    but since <v1,v1> is not 0 (because v1 is not 0), we must have a = 0.

    now show that <v2,u> = 0 implies b = 0, and that <v3,u> = 0 implies c = 0, showing {v1,v2,v3} is linearly independent.

    the only assumption we make is that none of the v's are 0-vectors, so any orthogonal set of non-zero vectors in R^3 is linearly independent.
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  6. #6
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    Re: Orthogonal sets

    Quote Originally Posted by Deveno View Post
    sure, just add the 0-vector [0,0,0].

    the point being, if we have an orthogonal set in R^3 that is all non-zero vectors, say v1,v2,v3. then consider any linear combination:

    u = av1 + bv2 + cv3. if this is 0, then <v1,u> = 0, since <v1,0> = 0 (here <x,y> is just the dot product in R^3).

    but <v1,u> = <v1,av1> + <v1,bv2> + <v1,cv3> = a<v1,v1> + b<v1,v2> + c<v1,v3> = a<v1,v1> (by orthogonality).

    but since <v1,v1> is not 0 (because v1 is not 0), we must have a = 0.

    now show that <v2,u> = 0 implies b = 0, and that <v3,u> = 0 implies c = 0, showing {v1,v2,v3} is linearly independent.

    the only assumption we make is that none of the v's are 0-vectors, so any orthogonal set of non-zero vectors in R^3 is linearly independent.
    Thanks a lot, this helps a ton. One more question. Does adding the zero vector still make the set orthogonal?
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  7. #7
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    Re: Orthogonal sets

    yes, because the 0-vector is orthogonal to every other vector.
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  8. #8
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    Re: Orthogonal sets

    Quote Originally Posted by Deveno View Post
    yes, because the 0-vector is orthogonal to every other vector.
    Got it, thanks again.
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