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In my current lecture I regularly encounter usage of the real part of, say, a scalar product of two vectors similar to angles in classical geometry. For example in Hilbert space theory:

Let $H$ be a Hilbert space, $C \subset H$ be convex and closed. For $x_0 \in H$, $x \in C$ is the best approximation of $x_0$ by $C$, iff $\forall y \in C : Re \langle x - x_0, y - x_0 \rangle \leq 0$

As much as it is intuitive (and the proof itself is no problem), I do not know how to interpret this. So the question (not necessarily connected to the above example theorem) is

Is there any useful interpretation to the real part of a scalar product in complex vector spaces?

if there is none, in best case the real part is just used for convinience, and the author of my book wants to grasp a more general concept. - in worst case, not such interpretation exists.

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Don't know if this is relevant, but is you have a complex inner product space $\mathbf{V}$, and you consider $\mathbf{V}$ as a real vector space by restriction of scalars, then $[x,y]=\mathrm{Re}\langle x,y\rangle$ gives a real inner product for $\mathbf{V}$ (with the additional property that $[x,ix]=0$ for all $x\in\mathbf{V}$, so the "complex copy" is orthogonal to the "real" copy). –  Arturo Magidin Mar 27 '11 at 4:18
    
What @Arturo said is the key idea. The 1 dimensional case will make it clear what's going on: you can think of $\mathbb C$ as a 1 dimensional complex vector space, or as the two real dimensional $\mathbb R^2$. By taking the real part of the Hermitian inner product, you obtain the real inner product on $\mathbb R^2$, and the linear operator that corresponds to multiplication by $i$ is now orthogonal. (A quick remark: the imaginary part of this inner product is then the standard symplectic form.) –  Sam Lisi Mar 27 '11 at 13:22
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I believe it is a notion similar to that of angle in real spaces, but you can't quite call it so. However, I've seen some authors do that.

The relationship between angles and lengths in real vector spaces is given by the formula

$$\lVert v+w \rVert^2 = \lVert v \rVert^2 + \lVert w \rVert^2 + 2\langle v, w\rangle.$$

Indeed, if we define

$$\cos \theta_{v, w} \lVert v \rVert \lVert w \rVert= \langle v, w \rangle,$$

we recover the familiar law of cosines

$$\lVert v+w \rVert^2=\lVert v \rVert^2 + \lVert w \rVert^2 +2\cos \theta_{v, w} \lVert v\rVert \lVert w \rVert.$$

This last formula holds true in a complex vector space if we define

$$\cos \theta_{v, w} \lVert v \rVert \lVert w \rVert= \text{Re} \langle v, w \rangle.$$

This is one reason why we might call "angle" this number $\theta_{v,w}$. Another reason is given by Arturo's observation above: for example, the formula

$$ \text{Re} (v \overline{w}), \quad v, w \in \mathbb{C}$$

gives us the real dot product if $\mathbb{C}$ is identified to the Euclidean plane.

However, there are drawbacks: the most striking (IMHO) is that parallel vectors can have zero angle. Take any non null $v$ and put $w=i v$. Then $\text{Re}\langle v, w \rangle=0$, but would you call them orthogonal?

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v and w are not parallel, but have a 90 degree angle from the coefficient i. Thus it is ok to call them orthogonal.

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