Triangle inequality proof

**Danneedshelp** · July 25th 2009, 08:25 PM

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.

**o_O** · July 25th 2009, 08:56 PM

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:

$\begin{aligned} \left| \vec{u} + \vec{v}\right| & = \sqrt{\left(kv_1 + v_1\right)^2 + \left(kv_2 + v_2\right)^2 + \cdots + \left(kv_n + v_n\right)^2} \\ & = \sqrt{ \left[(k+1)v_1\right]^2 + \left[(k+1)v_2\right]^2 + \cdots + \left[(k+1)v_n\right]^2 } \\ & \ \vdots \\ & = k|v| + |v| \end{aligned}$

**Moo** · July 26th 2009, 12:00 AM

Hello,

Originally Posted by o_O

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,

Originally Posted by Danneedshelp

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.

**Opalg** · July 26th 2009, 02:12 AM

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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