9 is not like 8

Number Theory Level 4

It is fascinating to study $n\times n$ matrices with integer entries whose row vectors are orthogonal and all have the same magnitude, $m$ (the column vectors will then be orthogonal and have magnitude $m$ as well).

For example, consider $A=\begin{bmatrix}1 & 2\\ -2 & 1\end{bmatrix}$ with $m=\sqrt{5},$ or $B=\begin{bmatrix}2 & 2 & 1\\ -2 & 1 &2\\ 1 & -2 & 2\end{bmatrix}$ with $m=3.$

In particular, it is interesting to investigate which values of $m$ are possible for a given size of the matrix. Let's explore this issue with two True/False questions.

True or False?

(I) In the case $n=9,$ the possible values of $m$ are all non-negative integers.
(II) In the case $n=8,$ the possible values of $m$ are the square roots of all non-negative integers.

Bonus Question: Do all odd numbers "behave like" 9 in (I), and do all even numbers behave like 8 in (II)?

(I) is true but (II) is false (II) is true but (I) is false Both (I) and (II) are true Both (I) and (II) are false

1 solution

Otto Bretscher
Oct 29, 2018

If $n$ is odd (for example, $n=9$ ) and $A$ is of the required form, having orthogonal rows of the same magnitude, $m$ , then $AA^T=qI_n$ where $q=m^2\in \mathbb{Z}$ . Thus $(\det A)^2=q^n$ and $|\det A|=q^{n/2}$ .Since $q^{n/2}$ is an integer, $q$ will be a perfect square, so that $m=\sqrt{q}$ is an integer, as claimed.

If $n=4$ and $q$ is any positive integer, then we can write $q=a^2+b^2+c^2+d^2$ for integers $a,b,c,d$ , by Lagrange's four-square theorem. Consider the matrix

$B=\begin{bmatrix} a&b&c&d\\-b&a&-d&c\\-c&d&a&-b\\-d&-c&b&a \end{bmatrix}$

that can be used to define the quaternion $a+bi+cj+dk$ . Note that $B$ has orthogonal rows with a common magnitude of $m=\sqrt{q}$ .

If $n=4m$ (for example, $n=8$ ), we can form a matrix $A$ that has $m$ $B$ 's along the diagonal and 0's elsewhere.

9 is not like 8

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