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Math Help - Triangle inequality proof

  1. #1
    Senior Member Danneedshelp's Avatar
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    Triangle inequality proof

    Q: Prove that ||\vec{v}+\vec{u}||=||\vec{u}||+||\vec{v}|| if and only if \vec{u} and \vec{v} have the same direction.

    A: Well, if \vec{u} the same direction of \vec{v}, then \vec{u} must be a multiple of \vec{v}. Thus, \vec{u}=k\vec{v} where k\in{\mathbb{R}} (I am not sure if I lost any generality by choosing \vec{u}). From here I expanded ||\vec{v}+\vec{u}|| and ended up with some messy algebra and not the result I was looking for.

    In the other direction I started with ||\vec{v}+\vec{u}||=... and worked my way down to =||\vec{u}||^{2}+|2<\vec{u},\vec{v}>|+||\vec{v}||^  {2} and figured I have to introduce the Cauchy-Schawrz Ineguality here, but I am not sure how.

    I am having a hard time formalizing this proof. All I have is a bunch of dead und scratch work. It seems that this proof should flow with ease, because, geometrically, it is pretty obviouse.

    I would appreciate some guidence.

    Thank you.
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  2. #2
    o_O
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    Let \vec{u} = \left<u_1, u_2, \cdots, u_n\right> and \vec{v} = \left<v_1, v_2, \cdots, v_n\right>.

    As you noted, \vec{u} = k\vec{v} and so:

     \begin{aligned} \left| \vec{u} + \vec{v}\right|  & = \sqrt{\left(kv_1 + v_1\right)^2 + \left(kv_2 + v_2\right)^2 + \cdots + \left(kv_n + v_n\right)^2} \\ & = \sqrt{ \left[(k+1)v_1\right]^2 + \left[(k+1)v_2\right]^2 + \cdots + \left[(k+1)v_n\right]^2 } \\ & \ \vdots \\ & = k|v| + |v|  \end{aligned}
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  3. #3
    Moo
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    Hello,
    Quote Originally Posted by o_O View Post
    Let \vec{u} = \left<u_1, u_2, \cdots, u_n\right> and \vec{v} = \left<v_1, v_2, \cdots, v_n\right>.

    As you noted, \vec{u} = k\vec{v} and so:

     \begin{aligned} \left| \vec{u} + \vec{v}\right|  & = \sqrt{\left(kv_1 + v_1\right)^2 + \left(kv_2 + v_2\right)^2 + \cdots + \left(kv_n + v_n\right)^2} \\ & = \sqrt{ \left[(k+1)v_1\right]^2 + \left[(k+1)v_2\right]^2 + \cdots + \left[(k+1)v_n\right]^2 } \\ & \ \vdots \\ & = k|v| + |v|  \end{aligned}
    Are you sure you can take wlog this norm ? Though all norms are equivalent in finite dimensions, we're talking about a norm in general, here.
    What you didn't say, and that is helpful for the solution, is that k has to be positive.

    Otherwise,
    Quote Originally Posted by Danneedshelp View Post
    Q: Prove that ||\vec{v}+\vec{u}||=||\vec{u}||+||\vec{v}|| if and only if \vec{u} and \vec{v} have the same direction.
    They wouldn't have the same direction as they would go in opposit directions.

    A: Well, if \vec{u} the same direction of \vec{v}, then \vec{u} must be a multiple of \vec{v}. Thus, \vec{u}=k\vec{v} where k\in{\mathbb{R}} (I am not sure if I lost any generality by choosing \vec{u}).
    It's okay, there is no loss of generality. You choose u because it's the assumption of the implication.

    From here I expanded ||\vec{v}+\vec{u}|| and ended up with some messy algebra and not the result I was looking for.
    Actually, you just have to use the axioms/definition of a norm.

    There is the first axiom (positive homogeneity) : \|a \vec{u}\|=|a| ~\|\vec{u}\|, where a is a real number.

    So in this case (I'll forget the \vec...too long ^^), \|u+v\|=\|kv+v\|=(k+1)~ \|v\|=k \|v\|+\|v\|=\|kv\|+\|v\|=\|u\|+\|v\|

    In the other direction I started with ||\vec{v}+\vec{u}||=... and worked my way down to =||\vec{u}||^{2}+|2<\vec{u},\vec{v}>|+||\vec{v}||^  {2} and figured I have to introduce the Cauchy-Schawrz Ineguality here, but I am not sure how.

    I am having a hard time formalizing this proof. All I have is a bunch of dead und scratch work. It seems that this proof should flow with ease, because, geometrically, it is pretty obviouse.

    I would appreciate some guidence.

    Thank you.
    A scalar product gives a norm.
    So let's define <,> as the scalar product associated with this norm.

    Now, we assume that \|u+v\|=\|u\|+\|v\| and we want to prove that there exists k\in\mathbb{R}_+ such that u=kv

    \|u+v\|^2=\langle u+v,u+v\rangle =\dots=\|u\|^2+\|v\|^2+2\langle u,v\rangle by definition.

    Now, we know that \|u+v\|=\|u\|+\|v\| \Rightarrow \|u+v\|^2=\|u\|^2+\|v\|^2+2\|u\|\|v\|

    So \langle u,v\rangle=\|u\|\|v\|

    And here, apply Cauchy-Schwarz inequality, stating that there is equality if and only if u and v are linearly dependent.
    That is to say u=kv, but where k is not necessarily positive.

    However, the equivalence only stands for k positive.
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  4. #4
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    Here's a little trick that avoids using Cauchy–Schwarz. You have already seen that if \|u+v\| = \|u\|+\|v\| then \langle u,v\rangle = \|u\|\|v\|. Let w = \|v\|u-\|u\|v. Then \|w\|^2 = \langle w,w\rangle = \langle \|v\|u-\|u\|v,\|v\|u-\|u\|v\rangle, which simplifies to 2\|u\|v\|(\|u\|\|v\| - \langle u,v\rangle). If \langle u,v\rangle = \|u\|\|v\| then it follows that w=0, from which v = \tfrac{\|v\|}{\|u\|}u.
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