Is there a way of getting these vectors linearly independent?

Question

I'm hoping that we can find a way to get the following matrix to have full rank:

$$ \begin{bmatrix} (x^a)_1 & (x^{2a})_1 & (x^{3a})_1 & \dots & (x^{na})_1\\ (x^a)_2 & (x^{2a})_2 & (x^{3a})_2 & \dots & (x^{na})_2\\ (x^a)_3 & (x^{2a})_3 & (x^{3a})_3 & \dots & (x^{na})_3\\ \dots & \dots & \dots & \dots & \dots \\ (x^a)_m & (x^{2a})_m & (x^{3a})_m & \dots & (x^{na})_m\\ \end{bmatrix} $$

Here $x$ has multiplicative order $na, n \in \mathbb{N}, a \in \mathbb{N}$. The subscripts $1,2,3,\dots, m$ denote the $m$th component in an $m$ component number system. For example, if we're working with quaternions, then $m=4$, and $x_1$ is the real component, $x_2$ the $i$ component, $x_3$ the $j$ component, and $x_4$ the $k$ component.

Can we find an $x$ with multiplicative order $na$ that will give the matrix full rank?

Answer 1

A few initial observations:

1. Using modular arithmetic can only hurt us, because it makes more numbers equal to other numbers, so that it becomes more likely that a row is a linear combination of other rows.
2. If by "$m$ component number system" you mean something where the components are real numbers, multiplication is associative, and division is possible, then the only number systems like that are the reals, the complexes, and the quaternions.
3. Once we have an example for the $n=m$ case, then if it weren't for the order condition, we could keep the same $x$ and just throw in more columns for $n>m$ because the row rank can't go down, and it's already full with only $m$ rows.
4. In the complex numbers and in the quaternions you can take the $a^\text{th}$ root of numbers without any problems, so if it weren't for the "order $na$" condition, the $x^a$ could be replaced by $x$ in those cases without losing any generality.

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The real numbers ($m=1$):

If it weren't for the order condition, then any nonzero $x$ works for any $n,a$, and $x=0$ works for no $n,a\ge1$.

To take the order condition into account, note that the only real numbers with finite order are $\pm1$, so the only allowed cases are $x=-1,a=2,n=1$, $x=-1,a=1,n=2$, and $x=1,a=1,n=1$. $\left[(-1)^2\right]=[1]$, $\begin{bmatrix}-1&1\end{bmatrix}$, and $[1]$ all have full rank.

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The complex numbers ($m=2$):

First, let's ignore the order condition and figure out what solutions we can get with $a=1,n=2$. Since the matrix is square, we can simply use the determinant to test for full rank. If $x=A+Bi$, then $x^2=A^2-B^2+2ABi$, so that the matrix is $\begin{bmatrix}A&A^2-B^2\\B&2AB\end{bmatrix}$. The determinant is $B(A^2+B^2)$ which is nonzero exactly when $B\ne0$. Note that there is no hope of a $B=0$ case having full rank even because of a later column ($n>2$), because if $B=0$ then the second row contains all zeros.

Now, when we impose the order condition, we need the order to be $na$ and we need $x^a$ not to be real. The only finite order real numbers are $\pm1$ and $n\ge2$ so the only problem to worry about is $x^a=-1$, $n=2$. But conversely, order $2a\,$ forces $x^a=-1$ since $x^a$ is a square root of $1$ and the order isn't $a$. Therefore, just pick $n>2$ and any $a$ and let $x$ be a primitive $\left(na\right)^{\text{th}}$ root of unity, and it'll work (and nothing else will).

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The quaternions ($m=4$):

Again, let's ignore the order condition and assume $a=1,n=4$ this time. Letting $x=R+Ai+Bj+Ck$, our matrix becomes $$\begin{bmatrix}R&R^2-A^2-B^2-C^2&R\left(R^2-3A^2-3B^2-3C^2\right)&\star\\A&2AR&A(3R^2-A^2-B^2-C^2)&4AR(R^2-A^2-B^2-C^2)\\B&2BR&B(3R^2-A^2-B^2-C^2)&4BR(R^2-A^2-B^2-C^2)\\C&2CR&C(3R^2-A^2-B^2-C^2)&4CR(R^2-A^2-B^2-C^2)\end{bmatrix}$$ Where $\star=R^4+A^4+B^4+C^4-6R^2(A^2+B^2+C^2)+2\left(A^2B^2+A^2C^2+B^2C^2\right)$. The problem with this is that rows 2,3, and 4 are all multiples of each other (or worse: zero), so this $4\times4$ matrix can have rank at most 2. Like in the complex case, the rank equals 2 as long as $x$ isn't real.

There is also no hope of a higher rank matrix by taking $n>4$ (you can prove this with the polar form of a quaternion, for instance). This is all related to the fact that the commutative subrings of the quaternions are copies of $\mathbb C$.

We don't even have to think about the order condition because there's no way to get full rank, period.