We have the scalar product $\left \langle \cdot, \cdot \right \rangle$. How is its belonging norm $\left \| \cdot \right \|$ defined?

Question

I'm not sure if I got it correctly because these dots are confusing. They just stand for "any", like variables $x,y$ as example?

If so I think they are just asking for the axioms of a norm?

Let $V$ be a $K$ vector space for $\left\{\mathbb{R}, \mathbb{C}\right\}$. A non-negative real function $\left \| \cdot \right \|: V \rightarrow [0, \infty )$ is called norm, if it has following properties:

positivity: $\left \| a \right \| \geq 0 \forall a \in V$ and $\left \| a \right \| >0 \forall a \in V$ with $a \neq 0$
homogeneity: $\left \| ka \right \| = |k| \cdot \left \| a \right \| \forall a \in V, k \in K$
triangle inequality: $\left \| a+b \right \| \leq \left \| a \right \|+\left \| b \right \| \forall a,b \in V$

Edit: Or would this even be better, and shorter:

Let $V$ be a $K$ vector space with a scalar product $\left \langle \cdot, \cdot \right \rangle$ whereby $K \in \left\{R, C\right\}$. Then there is a norm $\left \| \cdot \right \|$ on $V$ defined by

$$\left \| a \right \| = \sqrt{\left \langle a,a \right \rangle}, a \in V$$

Is that what was asked for? Please help me, this is not homework.

Yes, the dots are placeholders for variables. The norm ||x|| is just the inner product <x, x>. — Amey Joshi, Mar 17 '17 at 11:37
I think you could start with the 5 properties of inner product, and arrive at it being a norm — Jay Zha, Mar 17 '17 at 11:39

Scientifica · Accepted Answer · 2017-03-17T11:39:25.110

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Yes. The dots play kind of the role of a placeholder. We simply write $\|\cdot\|$ instead of \begin{align*} V&\to\mathbb{R}\\ v&\mapsto \|v\| \end{align*}

and $\langle\cdot,\cdot\rangle$ instead of \begin{align*} V\times V&\to\mathbb{R}\\ (u,v)&\mapsto\langle u,v\rangle \end{align*}

Whenever $\langle\cdot,\cdot\rangle$ is an inner product on a vector space $V$, the norm $\|\cdot\|$ on $V$ induced by the inner product $\langle\cdot,\cdot\rangle$ is defined by:

\begin{align*} V&\to\mathbb{R}\\ v&\mapsto\|v\|=\sqrt{\langle v,v\rangle} \end{align*}

Now, as you said, you should show that the "norm" we just defined is indeed a norm on $V$, which means that it satisfies the 3 properties that you gave.

edited Mar 17 '17 at 11:39

answered Mar 17 '17 at 11:37

Scientifica

8,781

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Maybe a square root missing? – GaC Mar 17 '17 at 11:39
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@SpettroDiA Oh my bad! Yes, you're right! – Scientifica Mar 17 '17 at 11:39
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Is my edit alright now, does it answer the task? Tried to use your answer for it. – tenepolis Mar 17 '17 at 11:40
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@tenepolis Yes that's right. Just a better way to say it: "the map $v\mapsto\sqrt{\langle v,v\rangle}$ is a norm on $V$. – Scientifica Mar 17 '17 at 11:42
Well, the three conditions mean that this map is indeed a norm. The first part means that this isn't any arbitrary norm: it is a norm associated to the inner product on your space in a nice way, for example you can prove that for this norm: $|u+v|^2=|u|^2+|v|^2+2\Re{\langle u,v\rangle}$ (where $\Re$ denoted the real part; just a precaution if you're working on a complex vector pace). Another arbitrary norm may not relate to your inner product like that... – Scientifica Mar 17 '17 at 11:51
... In fact, there are some inner products that are not induced by any norm. – Scientifica Mar 17 '17 at 11:51
@tenepolis You didn't accept my answer. This normally means that you're not satisfied by my answer, or that it is incomplete. If that's the case, would you tell me what do you still not understand? – Scientifica Mar 17 '17 at 19:19
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Actually I did accept it, but I guess there are some bugs on phone or something. Accepted on computer now ^^ – tenepolis Mar 17 '17 at 19:20
@tenepolis Oh I see. Glad it was useful :) – Scientifica Mar 17 '17 at 19:25

We have the scalar product $\left \langle \cdot, \cdot \right \rangle$. How is its belonging norm $\left \| \cdot \right \|$ defined?

1 Answers1