Do all angles occur in Hilbert spaces?

Let $X$ be a Hilbert space with scalar product $(\cdot,\cdot)$. Then for two vectors $v,w$ of norm $1$, we can interpret $(v,w)$ as an angle, so that $(v,w)=\cos(\varphi)$ for a unique angle $\varphi\in[0,\pi)$.

My question is the following: Let $\varphi'\in[0,\pi)$ and $v'\in X$ with $\|v'\|=1$ (norm induced by the scalar product) be given. Is there a vector $w'\in X$ with $(v',w')=\cos(\varphi')$?

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Did you want to write that there is a unique angle in $[0,\pi)$ instead of $[0,2\pi)$? – Martin Sleziak Aug 1 '12 at 10:50
For this to work, the Hilbert space must have dimension at least 2. – GEdgar Aug 1 '12 at 12:09
Yeah, I meant $[0,\pi)$, sorry. I edited the question. – Sh4pe Aug 2 '12 at 8:50

Let $w$ be orthogonal to $v$, $\lVert v \rVert = \lVert w \rVert = 1$. We have $$\lVert \sin(\phi) w + \cos(\phi) v \rVert^2 = \sin(\phi)^2 + \cos(\phi)^2 = 1$$ and $$(v, \sin(\phi) w + \cos(\phi) v) = \cos(\phi)$$

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Great, thank you! – Sh4pe Aug 2 '12 at 8:49

Every Hilbert space has orthonormal basis. The basis can be chosen is such way that it contains the given vector $v'$.

If the basis has at least two elements (=if the space $X$ is not 1-dimensional), then you have two orthogonal vectors $v'$ and $w$ in the basis. Now if $w'=v'+cw$ for some $c\in\mathbb R$, then $$(v',w')=(v',cv'+w)=c(v',v')=c,$$ and we have $$\cos\varphi=\frac{(v',w')}{\|v'\|\cdot\|w'\|}=\frac{c}{\sqrt{1+c^2}}.$$ The expression $c/\sqrt{1+c^2}$ can attain any value in $(-1,1)$.

We can obtain $\cos\varphi=1$ by choosing $w'=v'$ and $\cos\varphi=-1$ by choosing $w'=-v'$.

So you can obtain arbitrary value of the angle between $v'$ and $w'$.

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