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Math Help - Linear operator proof

  1. #1
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    Linear operator proof

    How do we prove this:


    Btw, here 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)

    So, I think to show that a mapping is linear we must show that it satisfies the additivity and homogeneity conditions. But I don't know how to prove these conditions here but here's what I think so far

    So, for vectors x_1,x_2 and scalars a and b, T(x)= <x,u> v becomes

    T(ax_1+bx_2)= \left\langle ax_1+bx_2,u \right\rangle v

    By the additivity property we get

    = (\left\langle ax_1,u \right\rangle + \left\langle bx_2,u \right\rangle )v

    I don't know what to do after this point. Can anyone help?
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  2. #2
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    Quote Originally Posted by demode View Post
    How do we prove this:


    Btw, here 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)

    So, I think to show that a mapping is linear we must show that it satisfies the additivity and homogeneity conditions. But I don't know how to prove these conditions here but here's what I think so far

    So, for vectors x_1,x_2 and scalars a and b, T(x)= <x,u> v becomes

    T(ax_1+bx_2)= \left\langle ax_1+bx_2,u \right\rangle v

    By the additivity property we get

    = (\left\langle ax_1,u \right\rangle + \left\langle bx_2,u \right\rangle )v

    I don't know what to do after this point. Can anyone help?
    To show that a transformation T:V \to W is linear, you need to show that for any scalar \alpha \in F and any two vectors v_1,v_2 \in V, T(\alpha v_1 + v_2) = \alpha T(v_1) + T(v_2).

    Now in your case, let \alpha be a scalar and x_1,x_2 be vectors in V. Then, T(\alpha x_1 + x_2) = <\alpha x_1 + x_2, u>v.

    Now, you need to show that T(\alpha x_1 + x_2) = \alpha T(x_1) + T(x_2). Note that \alpha T(x_1) + T(x_2) = \alpha <x_1,u>v ~ + <x_2,u>v. Can you show the equality now?
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  3. #3
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    Quote Originally Posted by Defunkt View Post
    To show that a transformation T:V \to W is linear, you need to show that for any scalar \alpha \in F and any two vectors v_1,v_2 \in V, T(\alpha v_1 + v_2) = \alpha T(v_1) + T(v_2).

    Now in your case, let \alpha be a scalar and x_1,x_2 be vectors in V. Then, T(\alpha x_1 + x_2) = <\alpha x_1 + x_2, u>v.

    Now, you need to show that T(\alpha x_1 + x_2) = \alpha T(x_1) + T(x_2). Note that \alpha T(x_1) + T(x_2) = \alpha <x_1,u>v ~ + <x_2,u>v. Can you show the equality now?
    T(\alpha x_1+x_2) = \left\langle \alpha x_1+x_2,u \right\rangle v

    By the homogeneity property:

    =\alpha \left\langle x_1+x_2,u \right\rangle v

    By the additivity property:

    \alpha ( \left\langle x_1,u \right\rangle + \left\langle x_2,u \right\rangle)v

    Now what do I need to do? Do I just distribute the "v"?
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  4. #4
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    Quote Originally Posted by demode View Post
    T(\alpha x_1+x_2) = \left\langle \alpha x_1+x_2,u \right\rangle v

    By the homogeneity property:

    =\alpha \left\langle x_1+x_2,u \right\rangle v

    This part is not correct. You know that <\alpha v_1, v_2> = \alpha<v_1, v_2> but this is not the case here. You need to use additivity first.

    By the additivity property:

    \alpha ( \left\langle x_1,u \right\rangle + \left\langle x_2,u \right\rangle)v

    Now what do I need to do? Do I just distribute the "v"?
    .
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  5. #5
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    Is the following right?

    T(\alpha x_1,x_2)= \left\langle \alpha x_1 + x_2, u \right\rangle v

    = ( \left\langle \alpha x_1,u \right\rangle + \left\langle x_2,u \right\rangle )v

    = \left\langle \alpha x_1,u \right\rangle v+ \left\langle x_2,u \right\rangle v

    = \alpha \left\langle x_1,u \right\rangle v + \left\langle x_2,u \right\rangle v
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  6. #6
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    Quote Originally Posted by demode View Post
    Is the following right?

    T(\alpha x_1,x_2)= \left\langle \alpha x_1 + x_2, u \right\rangle v

    = ( \left\langle \alpha x_1,u \right\rangle + \left\langle x_2,u \right\rangle )v

    = \left\langle \alpha x_1,u \right\rangle v+ \left\langle x_2,u \right\rangle v

    = \alpha \left\langle x_1,u \right\rangle v + \left\langle x_2,u \right\rangle v
    Yes, that is correct, and it is exact the equality you were asked to show.
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