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Math Help - inner product on R^2

  1. #1
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    inner product on R^2

    We'll define the following operation for vectors u and v in \mathbb{R}^2:

    <u,v>=u^TAv

    Where A= \left[\begin{array}{cc}1 & 2 \\ 2 & 4\end{array}\right]

    Which axioms of inner products are satisfied for <,>? Is it an inner product in \mathbb{R}^2?

    -----------------
    I'm not sure how to answer this question but I know that it must satisfy all the axioms in order to be an inner product. These are the properties of inner products:

    <u,v>=<v,u> (symmetry property)
    <u+v,w>=<u,w>+<v,w> (additivity property)
    <ku,v>=k<u,v> (homogeneity property)
    <v,v> ≥ 0 and <v,v>=0 if and only if v=0 (positivity property)

    I know that matrix multiplication not commutative, so I'm guessing the symmetry property is violated. I think the only condition it satisfies is the homogeneity. But I'm not sure how to go about proving the other ones. Any ideas?
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  2. #2
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    Quote Originally Posted by demode View Post
    We'll define the following operation for vectors u and v in \mathbb{R}^2:

    <u,v>=u^TAv

    Where A= \left[\begin{array}{cc}1 & 2 \\ 2 & 4\end{array}\right]

    Which axioms of inner products are satisfied for <,>? Is it an inner product in \mathbb{R}^2?

    -----------------
    I'm not sure how to answer this question but I know that it must satisfy all the axioms in order to be an inner product. These are the properties of inner products:

    <u,v>=<v,u> (symmetry property)
    <u+v,w>=<u,w>+<v,w> (additivity property)
    <ku,v>=k<u,v> (homogeneity property)
    <v,v> ≥ 0 and <v,v>=0 if and only if v=0 (positivity property)

    I know that matrix multiplication not commutative, so I'm guessing the symmetry property is violated. I think the only condition it satisfies is the homogeneity. But I'm not sure how to go about proving the other ones. Any ideas?

    Begin by checking positiveness, which you listed last but i's usually listed the first property an inner product must fulfill:

    v=\binom {x_1}{x_2}\,,\,\,v^tAv=(x_1\,\,x_2)\begin{pmatrix}  1&2\\2&4\end{pmatrix}\binom {x_1}{x_2}= (x_1\,\,x_2)\binom {x_1+2x_2}{2x_1+4x_2} =x_1^2+2x_1x_2+2x_1x_2+4x_2^2 =x_1^2+4x_1x_2+4x_2^2 = (x_1+2x_2)^2 ,

    and then you can see that it does NOT fulfill the condition (why?), so this is not an inner product (this was plain clear from the beginning if you knew about positive definite and semidefinite matrices but I assumed you don't know this yet).

    Tonio
    Last edited by tonio; March 27th 2010 at 01:47 PM. Reason: Typo fixed: NOW is in fact NOT. Thanx to Hallsofivy.
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  3. #3
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    Quote Originally Posted by tonio View Post
    Begin by checking positiveness, which you listed last but i's usually listed the first property an inner product must fulfill:

    v=\binom {x_1}{x_2}\,,\,\,v^tAv=(x_1\,\,x_2)\begin{pmatrix}  1&2\\2&4\end{pmatrix}\binom {x_1}{x_2}= (x_1\,\,x_2)\binom {x_1+2x_2}{2x_1+4x_2} =x_1^2+2x_1x_2+2x_1x_2+4x_2^2 =x_1^2+4x_1x_2+4x_2^2 = (x_1+2x_2)^2 ,

    and then you can see that it does NOW fulfill the condition
    Typo- Tonio clearly meant "it does NOT fulfill the condition"

    (why?), so this is not an inner product (this was plain clear from the beginning if you knew about positive definite and semidefinite matrices but I assumed you don't know this yet).

    Tonio
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    Thanks Tonio,

    if I may ask, what process did you use to get from (x_1\,\,x_2)\binom {x_1+2x_2}{2x_1+4x_2} to x_1^2+2x_1x_2+2x_1x_2+4x_2^2?
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  5. #5
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    Quote Originally Posted by demode View Post
    Thanks Tonio,

    if I may ask, what process did you use to get from (x_1\,\,x_2)\binom {x_1+2x_2}{2x_1+4x_2} to x_1^2+2x_1x_2+2x_1x_2+4x_2^2?
    It is common matrix multiplication of conformant matrices (which just means that they have appropriate dimensions to multiply by matrix multiplication).

    CB
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  6. #6
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    Quote Originally Posted by tonio View Post
    and then you can see that it does NOT fulfill the condition (why?), so this is not an inner product (this was plain clear from the beginning if you knew about positive definite and semidefinite matrices but I assumed you don't know this yet).

    Tonio
    Why?? As I see it (x_1+2x_2)^2 fulfill the positivity condition because (x_1+2x_2)^2 \geq 0 \forall x_1, x_2 and (x_1+2x_2)^2 = 0 if the vectors are zero so x_1, x_2 are zero.

    Now, can you check my work for the homogeneity property:

    Let k be any scalar.

    v=\binom {x_1}{x_2}\,,\,\,u=(y_1\,\,y_2)

    \left\langle ku,v \right\rangle =ku^TAv=(ky_1 \,\, ky_2) \binom {x_1+2x_2}{2x_1+4x_2} \binom {x_1}{x_2}

    = (ky_1\,\,ky_2)\binom {x_1+2x_2}{2x_1+4x_2}

    =ky_1 (x_1+2x_2) + ky_2 (2x_1+4x_2)

    =ky_1x_1+ky_12x_2+ky_22x_1+ky_24x_2

    = k(y_1x_1+y_12x_2+y_22x_1+y_24x_2) = k \left\langle u,v \right\rangle

    So it satisfies the homogeneity condition
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  7. #7
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    Quote Originally Posted by demode View Post
    Why?? As I see it (x_1+2x_2)^2 fulfill the positivity condition because (x_1+2x_2)^2 \geq 0 \forall x_1, x_2 and (x_1+2x_2)^2 = 0 if the vectors are zero so x_1, x_2 are zero.


    Not really: try the vector (-2,1) ...

    Tonio


    Now, can you check my work for the homogeneity property:

    Let k be any scalar.

    v=\binom {x_1}{x_2}\,,\,\,u=(y_1\,\,y_2)

    \left\langle ku,v \right\rangle =ku^TAv=(ky_1 \,\, ky_2) \binom {x_1+2x_2}{2x_1+4x_2} \binom {x_1}{x_2}

    = (ky_1\,\,ky_2)\binom {x_1+2x_2}{2x_1+4x_2}

    =ky_1 (x_1+2x_2) + ky_2 (2x_1+4x_2)

    =ky_1x_1+ky_12x_2+ky_22x_1+ky_24x_2

    = k(y_1x_1+y_12x_2+y_22x_1+y_24x_2) = k \left\langle u,v \right\rangle

    So it satisfies the homogeneity condition
    .
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    I see! Can you show me how the symmetry property can be proven? This is my last question.
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  9. #9
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    Quote Originally Posted by demode View Post
    I see! Can you show me how the symmetry property can be proven? This is my last question.

    Take two vectors u=(x_1,y_1)\,,\,v=(x_2,y_2)\in\mathbb{R}^2 and check that <u,v>=u^tAv=v^tAu=<v,u> . Not hard though slightly annoying work.

    Tonio
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    You may also note the fact that (v^tAu)^t = u^tAv however v^tAu \in \mathbb{R} and therefore its transpose is simply itself, that is:

    <v,u> = v^tAu = (v^tAu)^t = u^tAv = <u,v>.
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