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Suppose $H$ is a Hilbert space and the field we consider is $\mathbb{C}$, then given two arbitrary nonzero vectors $u$ and $v$ in $H$, how to define the angle between $u$ and $v$?

I tried to define it as $\cos^{-1} {\langle u, v\rangle}\over ||u|| \cdot||v||$, but $\langle u,v \rangle$ might be a compex number.

This question arose from a problem I want to solve. If the definition of angles does not exist, could you help me with the proof of the third claim below?

The original problem is:

Let $H$ be a Hilbert space, $M$ be a closed linear subspace of $H$ and $P$ be the orthogonal projection from $H$ to $M$. Let $x\in H$ and $a\in M$. Show that

  1. $|\langle x,a\rangle| \le ||Px||\cdot||a||$;
  2. if $a \in [[Px]]$, then $|\langle x,a\rangle|=||Px||\cdot||a||$;
  3. if $\mathbb{F} = \mathbb{C}$, then the angle between $x$ and $a$ is at least $\cos^{-1}({||Px||\over ||x||})$.

Proof of 1:

Since $P$ is the orthogonal projection from $H$ to $M$, we know $\text{im} P = M$, $\text{ker} P = M^{\perp}$ and $M \oplus M^{\perp} = H$. Thus, we have

$\begin{align}|\langle x, a \rangle| & = |\langle x - Px + Px, a \rangle| \\ & = |\langle x - Px, a\rangle + \langle Px, a\rangle| \\ & = |\langle Px,a \rangle | \quad \text{(by $x-Px \in M^{\perp}$ and $a \in M$) }\\ &\le ||Px||\cdot||a|| \quad \text{(by Cauchy-Schwarz inequality)}\end{align}$

Proof of 2:

If the proof of 1 we have proved that $|\langle x, a \rangle| = |\langle Px, a \rangle|$. Now, if $a \in [[Px]]$, then $\exists k \in \mathbb{F}$ s.t. $a = k Px$. Thus, $|\langle Px, a\rangle| = |\langle Px, kPx\rangle| = |k| \cdot|\langle Px,Px\rangle| = |k|\cdot||Px||\cdot ||Px|| = ||a|| \cdot||Px||$.

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  • $\begingroup$ What is the source of this problem? $\endgroup$ – Lord Shark the Unknown Mar 29 at 2:44
  • $\begingroup$ @LordSharktheUnknown The original problem is in Joseph Muscat's Functional Analysis: An introduction to Metric Spaces, Hilbert Spaces, and Banach Spaces p184 8.(a). published by Springer. But my professor substituted $\mathbb{R}$ with $\mathbb{C}$ and I don't know if it is reasonable. $\endgroup$ – U2647 Mar 29 at 2:54
  • $\begingroup$ Your professor didn't tell you their definition of angle? $\endgroup$ – Lord Shark the Unknown Mar 29 at 2:56
  • $\begingroup$ @LordSharktheUnknown No, he didn't. So I am wondering if there is a way to define angles in any Hilbert space with the field $\mathbb{C}$? If not, he might make a mistake or a typo. $\endgroup$ – U2647 Mar 29 at 3:13
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If $\langle \cdot,\cdot\rangle$ is a complex inner product on a complex vector space $H$, then its real part $(x,y)=\operatorname{Re}\langle x,y\rangle$ is a real inner product on $H$ (with the same induced norm as the complex inner product). So, angles can be defined using this real inner product, and this is normally what is meant by an angle in a complex inner product space.

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  • $\begingroup$ But what is the denominator? $\endgroup$ – U2647 Mar 29 at 3:18
  • $\begingroup$ There isn't any problem with the denominator...the formula you wrote in the question is correct if you just use the real inner product in place of $\langle\cdot,\cdot\rangle$. (Note that the real and complex inner products induce the same norm.) $\endgroup$ – Eric Wofsey Mar 29 at 3:42

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