1
$\begingroup$

Let $V$ be a vector space with positive-definite inner product $$\langle \cdot, \cdot \rangle: V \times V \to \mathbb{R}.$$ Let $\dim V = n$, $n \in \mathbb{Z}$. What is the largest number $m$, such that there exist $v_1, \dots, v_m \in V$ such that $\langle v_i, v_j\rangle < 0$ for all $i \neq j$?

I know that answer is $n + 1$ if $V$ is $\mathbb{R}^n$ with standard inner product. Here is the proof: https://mathoverflow.net/questions/31436/largest-number-of-vectors-with-pairwise-negative-dot-product

What if $V$ is arbitrary? For example, let $V$ be a space of polynomials with degree at most $k$ and define inner product by $$\langle p, q \rangle = \int_0^1 p(x)q(x)dx.$$

$\endgroup$
0
$\begingroup$

If you have a positive-definiteness condition on your inner product, then the answer is the same for any $n$-dimensional real vector space (with $n$ finite). You can see this by choosing an orthonormal basis for $V$ and mapping it to the standard basis in $\Bbb R^n$, extending this map linearly; the two are then isomorphic as inner product spaces.

If you relax positive-definiteness then the answer depends on the signature of the inner product, and can be infinite. For example, in Minkowski space you can have vectors such that $\langle v,v\rangle <0$. For such a $v$, every element of the set $\{\lambda v\}$ with $\lambda>0$ has negative inner product with every other element of the set.

$\endgroup$
  • $\begingroup$ Could you clarify? How can I map orthonormal basis of $V$ to standard basis of $R^n$? For example, orthonormal basis of space of polynomials with degree at most 2 is $e_1(x) = 1, e_2(x) = \sqrt{3}(-1 + 2x), e_3(x) = \sqrt{5}(1 - 6x + 6x^2)$. Also, I have positive-definiteness condition. I have edited my question. $\endgroup$ – edubrovskiy Nov 24 '15 at 19:43
  • $\begingroup$ Ok, I got it. So, I have an isomorphism $T: V \to \mathbb{R}^n$. I can choose $n + 1$ vectors with pairwise negative inner product in $\mathbb{R}^n$ and show that for each $u, v$ of these vectors $\langle T^{-1}u, T^{-1}v \rangle < 0$. $\endgroup$ – edubrovskiy Nov 24 '15 at 20:04
0
$\begingroup$

The space of polynomials is infinite-dimensional. So we can choose a sequence of polynomials $p_1,p_2,p_3, \ldots$ such that each $p_{n+1}$ is perpendicular to the span of $\{p_1,p_2, \ldots, p_n\}$. That gives an increasing chain of finite-dimensional subspaces $P_n = $ span$\{p_1,p_2, \ldots, p_n\}$.

By how we chose the polynomials $p_n$ we know the inner product restricted to each $P_n$ looks like the Euclidean inner product on $\mathbb R ^n$.

Then you can choose vectors $\{v_1,v_2, v_3, \ldots \} \subset V$ such that each $\{v_1,v_2, \ldots v_k, v_{k+1}\} \subset P_k$ form the vertex of a regular simplex at the origin. It follows the set $\{v_1,v_2, \ldots v_k, v_{k+1}\}$ has mutually negative dot products. By extension the same is true for $\{v_1,v_2, v_3, \ldots \}$.

Question: What if the space is finite-dimensional, but the inner product is not positive-definite?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.