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Let $\mathbb{F}_q$ be a finite field.

Prove that there is a mapping $\phi:\mathbb{F}_q\to\mathbb{R}^q$ such that:

  • For all $a\in\mathbb{F}_q$ such that $\lVert\phi(a)\rVert=1$.
  • For all $a,b\in\mathbb{F}_q$, $a\neq b$ and $\langle\phi(a),\phi(b)\rangle=-\frac{1}{q-1}$.

Note that $\langle\cdot, \cdot\rangle$ is an inner product over the field (dot product) and $\lVert\cdot\rVert$ is the induced norm.

Don Fanucci
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1 Answers1

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Let $\alpha=-\frac{1}{q-1}$ and $G$ be the $q \times q$ matrix given by $G_{ij}=\alpha$ if $i \ne j$ and $G_{ii}=1$.

We seek a set of $q$ vectors in $\mathbb R^q$ having $G$ as their Gram matrix.

Now, $G$ is a symmetric positive semidefinite matrix because its eigenvalues are $1-\alpha>0$ and $(q-1)\alpha+1=0$. Therefore, $G$ is the Gram matrix for some set of vectors.

Indeed, by the spectral theorem, $G=PDP^T=AA^T$ for $A=P \sqrt D$.

The rows of $A$ are the image of $\phi$.

lhf
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