Suppose we have some vector space $V$ for which we have defined an inner product $\langle \cdot\rangle$. Thus we have an inner product space. Is it true that $\forall x \in V : \lVert x\rVert := \sqrt{\langle x,x\rangle}$ is the only possible norm in $V$?

Let's consider the following example, that shows, what happens if it is not true (if by an inner product induced norm is not the only possible in this vector space):

Let $\mathcal{P}$ be a vector space of polynomials with an inner product defined as $$\langle P,Q\rangle=\int_0^1P(x)Q(x)\operatorname{d}x \quad \forall P,Q\in\mathcal{P}$$ Suppose we want to find an angle $\phi$ between two polynomials $P_1$ and $P_2$. We use a standard formulae for an angle between two vectors $$\cos\phi={\langle P_1,P_2\rangle \over \rVert P_1\lVert\rVert P_2\lVert}$$ So the norms in denominator should be the ones induced by the inner product, otherwise the angle will depend on how we define a norm.

So one more time: I suspect, that for a vector space with defined inner product, the norm of a vector, that lies in this vector space is also defined as induced by the inner product and is unambiguous. Is it true?

  • $\begingroup$ What about $|||x||| = C \cdot ||x||$ where $C > 0$ is a real? $\endgroup$ – William Jul 31 '14 at 11:20
  • $\begingroup$ different inner products establish different geometries , so orthogonality may changes.Orthogonality only makes sense with respect to a given inner product, not for an arbitrary vector space. $\endgroup$ – Finish Jul 31 '14 at 11:31

It will be possible to define multiple norms on any vector space (although they may be equivalent). For example, $||x|| = c\sqrt{\langle x , x \rangle }$ will also be a norm for any $c$. Indeed for your example, there is an infinite class of non-equivalent norms given by $$||P(x)||_p = \left ( \int_0^1\left |P(x)\right|^pdx\right )^\frac1p$$ for any $p \in \mathbb R, p \ge 1$.

However, in the definition of an angle as you have expressed it, the norm there must be the standard norm $||x|| = \sqrt{\langle x , x \rangle }$, and if we were to define a new norm, it would not necessarily be consistent with the inner product.

In short, when we work with an inner product space we usually assume that we're working with the standard norm. But that doesn't mean that there aren't other norms.

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  • $\begingroup$ So again, you mean that in the definition of an angle there stays the norm induced by inner product? Just by definition of an angle $\endgroup$ – Dmitry Kazakov Jul 31 '14 at 11:27
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    $\begingroup$ Exactly. Angles can only be defined in an inner product space. The norm in the definition of an angle is the induced norm $\endgroup$ – Mathmo123 Jul 31 '14 at 11:29
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    $\begingroup$ There's a general pattern. With an inner product space, we take the norm induced by the inner product unless explcitly stated otherwise. With a normed space, we take the metric induced by the norm unless explicitly stated otherwise. With a metric space, we take the topology induced by the metric unless explicitly stated otherwise. $\endgroup$ – Hagen von Eitzen Jul 31 '14 at 12:03

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