$\mathcal{H}$ is Hilbert space. A finite subtset of vectors $S = \{x_1, x_2, ..., x_n\}$ must have at least two identical vectors if:

Question

for any vector $\xi \in \mathcal{H}$, there exist two vectors $x_i, x_k$ in the set $S$ such that $\langle x_i, \xi \rangle = \langle x_k, \xi \rangle$. (two such vectors may depend on $\xi$)

If the set S satisfies the above property, are there two identical vectors in it?

I have been trying to construct a counterexample, but I didn't make it.

Moreover, is this question easy to solve when the Hilbert space is separable?

Answer 1

We may replace the finite set with the set of pairwise differences of the vectors, and turn the problem into the following:

$S=\{x_1, x_2, \cdots\}\subset\mathcal H$ be finite, such that $\forall \xi\in\mathcal H, \exists x_i\in S$ such that $(\xi, x_i)=0$.

Is it necessary that $0\in S$?

If $\mathcal H$ is infinite dimensional, then this is trivial, as we only need to pick any nonzero vector $\xi\in S^{\perp}$. Now we focus on the finite dimensional case.

We show that $S$ must contain the zero vector. If not, we may normalize $S$, i.e. assume $\|x_i\|=1$ for any $x_i\in S$. Let $\xi_1=x_1$, and we shall inductively construct $\xi_i$, such that $\xi$ has non-vanishing inner product with any of $x_1, x_2, \cdots, x_i$.

Assume we have constructed $\xi_1, \cdots, \xi_k$, if $(\xi_k, x_{k+1})\not=0$, then $\xi_{k+1}=\xi_k$ is fine. Otherwise, let $$\delta=\min_{i=1, 2, \cdots, k} |(\xi_k, x_i)|>0$$

Now $\xi_{k+1} = \xi_k + \frac{\delta}{2}x_{k+1}$ would do the job, as $|(\xi_{k+1}, x_i)|\ge|(\xi_k, x_i)| - \frac{\delta}{2}|(x_{k+1}, x_i)|\ge \frac{\delta}{2}$.

This is similar to what @Carl_Schildkraut did, but with an analysis flavor.

Answer 2

Yes, $S$ must contain two identical vectors.

Assume for the sake of contradiction that it doesn't, and let $T=\{x_i-x_j\colon 1\leq i<j\leq n\}$; note that $T$ is a finite set, $0\not\in T$, and for every $\xi\in\mathcal H$, there exists $v\in T$ with $\langle v,\xi\rangle=0$.

We'll show that, given any finite set $T\subset\mathcal H$ with $0\not\in T$, there exists some $v\in\mathcal H$ with $\langle v,\xi\rangle\neq 0$ for every $v$. Let $k$ be the smallest number for which, for every $\xi\in\mathcal H$, there are at least $k$ elements $v\in T$ with $\langle v,\xi\rangle=0$; we need to show that $k=0$. If $k$ is nonzero, take some vector $\xi_0\in\mathcal H$ for which $\langle v,\xi_0\rangle=0$ for exactly $k$ elements $v\in T$. Let one such $v$ be $v_0$, and, consider the vector $\xi=\xi_0+\lambda v_0$ for various values $\lambda$. As long as $\langle v_0,v\rangle=0$ or $\lambda\neq -\frac{\langle \xi_0,v\rangle}{\langle v_0,v\rangle}$ for every $v\in T\setminus U$, $\langle \xi,v\rangle$ will be nonzero for every $v\in T\setminus U$, since $$\langle \xi,v\rangle = \langle \xi_0,v\rangle+\lambda\langle v_0,v\rangle.$$ Also, as long as $\lambda\neq 0$, $\langle \xi,v_0\rangle=\lambda\langle v_0,v_0\rangle\neq 0$, since $\langle\cdot,\cdot\rangle$ is positive definite and $v_0\neq 0$. This means that $\langle\xi,v\rangle\neq 0$ for $v\in (T\setminus U)\cup \{v_0\}$, and so $\langle\xi,v\rangle=0$ for at most $k-1$ elements $v\in T$, contradicting the minimality of $k$. This occurs for all but finitely many values of $\lambda$, and so we can certainly choose some $\lambda$ for which it occurs, and gives a contradiction. Thus, $k$ must be $0$, as desired.

(One can also, if one wishes, phrase this algorithmically: start with some vector $\xi_0$ which may have $\langle\xi_0,v\rangle=0$ for lots of $v\in T$, transform it by adding a scalar multiple of one such $v$ into a vector which has $\langle\xi_1,v\rangle=0$ for a strictly smaller set of $v\in T$, and repeat.)

Note that this shows a much more general statement: $\mathcal H$ needs only be a vector space (over any field with at least $\binom n2+2$ elements, although I'm sure this too can be optimized) equipped with an inner product satisfying $\langle v,v\rangle\neq 0$ for all $v\neq 0$.