# Is it true that for an inner product space a norm of a vector is defined unambiguously?

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?

• What about $|||x||| = C \cdot ||x||$ where $C > 0$ is a real? – William Jul 31 '14 at 11:20
• 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. – 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.