Solving an equation in a Hilbert space

Question

Suppose $\mathbb{H}$ is a Hilbert space over the reals. Fix two vectors $\alpha$ and $\beta$ in $\mathbb{H}$. I came across a problem that reduces to the following: What are the possible units vector $h$ such that $\langle \alpha , h \rangle \langle \beta,h \rangle= \langle \alpha, \beta \rangle$ ? I could get the set of possible $h$ in the case that $\alpha=\beta$ and in the case that $\alpha$ and $\beta$ are orthogonal. Any idea with the more general formulation ?

Answer 1

WLOG we may assume that $\|\alpha\|=\|\beta\|=1$ and $\langle \alpha,\beta\rangle\ge 0.$ Then $\langle \alpha,\beta\rangle=\cos\theta$ for some $0\le \theta\le {\pi\over 2}.$ Let $V={\rm span}(\alpha,\beta).$ Any $h\in\mathcal{H}$ can be decomposed as $h=v+w,$ where $v\in V$ and $w\perp V.$ It suffices to find all possible $h\in V$ satisfying the condition. Then the general solution can be obtained by adding arbitrary elements in $V^\perp.$ We will skip the easy case when $\alpha=\beta.$ Otherwise the space $V$ is two dimensional if $\alpha\neq \beta.$ Let $h=x\alpha+y\beta.$ The condition is equivalent to $$(x+y\cos\theta ) (y+x\cos\theta )=\cos\theta$$ or equivalently to $$ xy(1+\cos^2\theta)=(1-x^2-y^2)\cos\theta$$ Thus we obtain a quadratic equation in two variables. For $\cos\theta =0$ we get $xy=0.$

Answer 2

To start, let's just acknowledge a special case: if $\alpha \perp \beta$, then the equation becomes $\langle \alpha, h \rangle \langle \beta, h \rangle = 0$. The set of unit vectors $h$ that satisfy this are the perpendicular vectors to at least one of $\alpha$ or $\beta$. So, we will assume $\alpha \not\perp \beta$. Note that this implies $\alpha \neq 0$ and $\beta \neq 0$, and that $\Bbb{H}$ is non-trivial.

Consider also the case where $\Bbb{H}$ is one-dimensional. It's not difficult to see that both unit vectors will work in this case, as $\alpha = ah$ and $\beta = bh$ for some $a, b \in \Bbb{R}$. So, let us assume the space has at least two dimensions.

The equation is equivalent to \begin{align*} \langle \alpha - \langle \alpha, h \rangle h, \beta \rangle = 0 &\iff \alpha - \langle \alpha, h \rangle h \perp \beta \\ &\iff P_{\{h\}^\perp}(\alpha) \perp \beta. \end{align*}

Note that, for any unit vector $h$, $\{h\}^\perp$ is a hyperplane through the origin, and every hyperplane corresponds to precisely two normal unit vectors $\pm h$. Also note that $\langle \alpha, h \rangle \langle \beta, h \rangle = \langle \alpha, \beta \rangle$ if and only if $\langle \alpha, -h \rangle \langle \beta, -h \rangle = \langle \alpha, \beta \rangle$. So, we can solve our problem by looking for hyperplanes $H$ through the origin in $\Bbb{H}$ where $P_H(\alpha) \perp \beta$, and find our solutions $h$ by taking the positive and negative unit normals of our solutions $H$.

Further, we can find all such hyperplanes by finding all the possible projections one can make of $\alpha$ onto a hyperplane. If $p$ is the projection of $\alpha$ onto hyperplane $H$ (through the origin) then we note that, if $p = \alpha$, then our condition $p \perp \beta$ would simplify to $\alpha \perp \beta$, which we assume not to be the case.

Thus, under our assumptions, $p \neq \alpha$, hence $p - \alpha$ is a normal direction to $H$. We obtain precisely two solutions to our original equation from $p$: $$h = \pm \frac{p - \alpha}{\|p - \alpha\|}.$$

Thus, we are interested in finding the complete set of possible projections of $\alpha$ onto a hyperplane through the origin. Let this set be $\mathcal{P}$. Using the above equation, the solution set to the original equation is parameterised by $\mathcal{P} \cap \{\beta\}^\perp$.

Let $\mathcal{P}'$ be the set of projections of $\alpha$ onto a line through the origin. I claim that $\mathcal{P} = \mathcal{P}'$.

Note that $0 \in \mathcal{P}$, as $\{\alpha\}^\perp$ is a hyperplane (recall $\alpha \neq 0$) onto which $\alpha$ projects to $0$. Any line in $\{\alpha\}^\perp$ (and one will exist, as $\Bbb{H}$ has dimension at least $2$) will also project $\alpha$ onto $0$, hence $0 \in \mathcal{P}'$.

If $x \in \mathcal{P} \setminus \{0\}$, then $L = \operatorname{span}\{x\}$ is a line onto which $\alpha$ projects onto $x$, hence $x \in \mathcal{P}'$. Conversely, if $x \in \mathcal{P}' \setminus \{0\}$, then $\{\alpha - x\}^\perp$ is a hyperplane onto which $\alpha$ projects onto $x$, so $x \in \mathcal{P}$. Thus, $\mathcal{P} = \mathcal{P}'$ as claimed. That is, $\mathcal{P}$ is the set of projections both onto hyperplanes, and onto lines, through the origin.

Let $$C := S_{\Bbb{H}}\left[\frac{\alpha}{2}; \frac{\|\alpha\|}{2}\right] = \left\{x \in \Bbb{H} : \left\|x - \frac{\alpha}{2}\right\| = \frac{\|\alpha\|}{2}\right\}.$$ I claim that $C = \mathcal{P}$. Note that: \begin{align*} \left\|x - \frac{\alpha}{2}\right\| = \frac{\|\alpha\|}{2} &\iff \left\|x - \frac{\alpha}{2}\right\|^2 = \frac{\|\alpha\|^2}{4} \\ &\iff \|x\|^2 - \langle x, \alpha \rangle + \frac{\|\alpha\|^2}{4} = \frac{\|\alpha\|^2}{4} \\ &\iff \|x\|^2 = \langle x, \alpha \rangle. \end{align*}

Suppose that $x \in C$. We know that $0 \in \mathcal{P}$ and $0 \in C$, so assume $x \neq 0$. Let $L = \operatorname{span} \{x\}$. Then, $$P_L(\alpha) = \frac{\langle \alpha, x \rangle}{\|x\|^2} x = x.$$ That is, $x \in \mathcal{P}' = \mathcal{P}$.

Conversely, suppose $x$ is the projection of $\alpha$ onto a line $L$ through the origin. Either $x = 0$, which lies in $C$, or $x \neq 0$. If $x \neq 0$, then $L = \operatorname{span}\{x\}$. In this case, we have $$x = P_L(\alpha) = \frac{\langle \alpha, x \rangle}{\|x\|^2} x,$$ and since $x \neq 0$, this implies $$\frac{\langle \alpha, x \rangle}{\|x\|^2} = 1 \implies x \in C.$$ Thus $\mathcal{P} = C$, as claimed. This gives us our family of solutions: $$h = \pm\frac{p - \alpha}{\|p - \alpha\|}, \qquad \text{where } p \in S_{\Bbb{H}}\left[\frac{\alpha}{2}; \frac{\|\alpha\|}{2}\right] \cap \{\beta\}^\perp,$$ where $\alpha \not\perp \beta$, and $$h \in (\{\alpha\}^\perp \cup \{\beta\}^\perp) \cap S_{\Bbb{H}}[0; 1]$$ when $\alpha \perp \beta$.

---

Some notes about this answer

I have checked this answer a few times, and it seems to be correct. However, I find it interesting that the solution is not obviously symmetric in $\alpha$ and $\beta$, even though the original equation clearly is. This gives me some pause; maybe the solution has some kind of error or hidden assumption that I'm missing?

I did some investigation in two dimensions, using Desmos, and it seems to validate the solution. I saw that the solution set is the same if you exchange $\alpha$ and $\beta$, and I visually saw the importance separating out the $\alpha \perp \beta$ case.

Unfortunately, two dimensions is not really sufficient to visualise the problem; I really need three dimensions for the picture to be truly convincing (the sphere intersected with a hyperplane is just two points in $\Bbb{R}^2$, and I would need to see a full circle at least). That said, I do stand by my solution, since the proof seems to be correct. It just would be nice to see an independent proof that this solution set is indeed symmetric.