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Equivalently, does every unit vector in $\mathbb{Q}^n$ belong to some orthonormal basis for $\mathbb{Q}^n$?

This is clearly true for $\mathbb{Q}^2$, and for $\mathbb{Q}^3$ it seems to be true for every vector I check. For example,

  • $\displaystyle\frac{1}{3}\!\begin{bmatrix}1 \\ 2 \\ 2\end{bmatrix}$ is the first column of $\displaystyle\frac{1}{3}\!\begin{bmatrix}1 & 2 & 2 \\ 2 & 1 & -2 \\ 2 & -2 & 1\end{bmatrix}$

  • $\displaystyle\frac{1}{7}\!\begin{bmatrix}2 \\ 3 \\ 6\end{bmatrix}$ is the first column of $\displaystyle\frac{1}{7}\!\begin{bmatrix}2 & 3 & 6 \\ 3 & -6 & 2 \\ 6 & 2 & -3\end{bmatrix}$

  • $\displaystyle\frac{1}{9}\!\begin{bmatrix}7 \\ 4 \\ 4\end{bmatrix}$ is the first column of $\displaystyle\frac{1}{9}\!\begin{bmatrix}7 & 4 & 4 \\ 4 & 1 & -8 \\ 4 & -8 & 1\end{bmatrix}$

Here's an example in $\mathbb{Q}^4$:

  • $\displaystyle\frac{1}{2}\!\begin{bmatrix}1 \\ 1 \\ 1 \\ 1\end{bmatrix}$ is the first column of $\displaystyle\frac{1}{2}\!\begin{bmatrix}1 & 1 & 1 & 1 \\ 1 & 1 & -1 & -1 \\ 1 & -1 & 1 & -1 \\ 1 & -1 & -1 & 1\end{bmatrix}$

Actually, all of the matrices in these examples are symmetric. This wouldn't be true if you permuted the entries of the vectors, but you could manage to make them some column of a symmetric orthogonal matrix. So we have the additional question:

Is every unit vector in $\mathbb{Q}^n$ a column of some rational symmetric orthogonal matrix?

Jim Belk
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1 Answers1

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Answer is yes.

In particular, if the first coordinate of the vector is not $-1$, the following matrix works:

enter image description here

This matrix represents the geometric reflection which sends your chosen vector to the standard unit vector.

If you want more information, you can look at Extending Orthonormal rational vectors.

Note: This result is generalized by Witt's Theorem.

Calvin Lin
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  • Fantastic! Incidentally, in the examples I gave, the least common denominator of the matrix entries is the same as the least common denominator of the original vector. Do you happen to know whether this can always be achieved? – Jim Belk Mar 29 '14 at 19:56
  • @JimBelk I have not looked into that aspect before. Why would that be interesting? – Calvin Lin Mar 29 '14 at 19:58
  • It says something interesting about integer matrices, i.e. that every integer vector with integer length is part of an orthogonal basis of integer vectors of that same length. – Jim Belk Mar 29 '14 at 19:59
  • For example, when $a_1 = 4/9$, $a_2 = 4/9$, and $a_3 = 7/9$, the matrix resulting from your formula has a least common denominator of $117$. – Jim Belk Mar 29 '14 at 20:10
  • @JimBelk Yes, I spoke too soon. I forgot about the $a_1 + 1$ factor in the denominator. When I though about this previously, I felt that the answer has to be yes, but didn't have proof. (Sorry, got confused with different versions of the question.) – Calvin Lin Mar 29 '14 at 20:13
  • @JimBelk Your question has been considered here. The best conclusion is that it's true for up to $n\leq 8$, and there is a counterexample for $n=9$ (Will Jaggy's 2nd answer). – Calvin Lin Mar 29 '14 at 20:27
  • Oh excellent! Thank you! – Jim Belk Mar 29 '14 at 20:31