Representing natural numbers as matrices by use of $\otimes$

Question

What I am wanting to do is to find a unique matrix representations for Natural numbers. Say I have the number $n$, how can I represent this number as a matrix in which I can do matrix multiplication on other natural number representations, say $m$ to get a matrix resulting in the actual number that would be the product i.e. a matrix representing $m*n$ I have found something akin to what I am wanting to do if I take the tensor product of 2 matrices For example, say I would like to represent the numbers $a*b$ and $c*d$ where $a,b,c,d \in \mathbb{N}$ and are prime : $$ \mbox{} \left[\begin{array} \\ a & 0 \\ 0 & b \end{array} \right] \otimes \mbox{} \left[\begin{array} \\ c & 0 \\ 0 & d \end{array} \right] = \mbox{} \left[\begin{array} \\ a*c & 0 & 0 & 0 \\ 0 & a*d & 0 & 0 \\ 0 & 0 & b*c & 0 \\ 0 & 0 & 0 & b*d \end{array} \right]$$ But as the product should be $a*b*c*d$, I would like to have the matrix representation: $$\mbox{} \left[\begin{array} \\ a & 0 & 0 & 0 \\ 0 & b & 0 & 0 \\ 0 & 0 & c & 0 \\ 0 & 0 & 0 & d \end{array} \right]$$

So basically I am just looking to represent each natural number uniquely in matrix format in which some operation gives me a new matrix which uniquely represents the product of natural numbers. I am hoping to do this with the tensor product as I would eventually like to represent numbers uniquely in a complex Hilbert space.

Thanks for your insight,

Brian

Answer 1

Let $\bf x$ vector contain natural numbers: ${\bf x}(a) = [x_1(a),x_2(a),\cdots]$ which is the multiplicity of respective prime in $a$'s prime factorization according to some enumeration of primes $[p_1,p_2,\cdots]$:

$$a = \prod_{\forall i} {(p_i)}^{x_i(a)}$$

Now you can build the matrix of outer product of standard basis elements with column vectors:$${\bf e_i} = \cases{1, \text{position }i\\0, \text{all other positions}}$$ $${\bf M}(a) = \sum_{\forall i} ({\bf e_i}{\bf e_i}^T)\cdot {(p_i)}^{{x_i}(a)}$$

Now a simple matrix product ${\bf M}(a) {\bf M}(b) = {\bf M}(ab)$ will have the same representation for $ab$ as it has for $a$ and $b$ and you don't have to use Kronecker product. Also note that $$\det({\bf M}({a})) = \prod_{\forall i}({p_i})^{x_i(a)}=a$$

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An alternative would be to store block diagonal matrix with entries: $\left[\begin{array}{cc}1&0\\x_i&1\end{array}\right]$. This way the matrix multiplication would give addition of the exponents which satisfies $$p^{x_i(a)} p^{x_i(b)} = p^{x_i(a)+x_i(b)}$$ One advantage of this representation is that we can be sure that we will get away with an integer matrix, even when doing division (although we must allow negative integers). This construction you can build with $${\bf M_2}(a) = {\bf I}_n \otimes \left[\begin{array}{cc}1&0\\0&1\end{array}\right] + \left(\sum_{\forall i} {\bf e_i} {\bf e_i}^T x_i(a) \right)\otimes \left[\begin{array}{cc}0&0\\1&0\end{array}\right]$$

So we do finally get an excuse to use the beloved $\otimes$ Kronecker product.

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Small examples: if we use $p = [2,3,5,\cdots]$ numbers $6$ and $9$ will have representation:

$${\bf M}(6) = \left[\begin{array}{ccc}2&0&0\\0&3&0\\0&0&1\end{array}\right], \hspace{1cm} {\bf M}(9) = \left[\begin{array}{ccc}1&0&0\\0&9&0\\0&0&1\end{array}\right]$$ And we can confirm that: $${\bf M}(6){\bf M}(9) = \left[\begin{array}{ccc}2&0&0\\0&27&0\\0&0&1\end{array}\right] = {\bf M}(54), \det({\bf M}(54)) = 2\times 27 = 54 = 6\times9$$

And the second type, where we for simplicity draw a grid to show the block structure imposed by the Kronecker product:

$${\bf M_2}(6) = \left[\begin{array}{cc|cc|cc}1&0&0&0&0&0\\\bf 1&1&0&0&0&0\\\hline0&0&1&0&0&0\\0&0&\bf 1&1&0&0\\\hline0&0&0&0&1&0\\0&0&0&0&\bf 0&1\end{array}\right],\hspace{1cm} {\bf M_2}(9) = \left[\begin{array}{cc|cc|cc}1&0&0&0&0&0\\\bf 0&1&0&0&0&0\\\hline0&0&1&0&0&0\\0&0&\bf 2&1&0&0\\\hline0&0&0&0&1&0\\0&0&0&0&\bf 0&1\end{array}\right]$$

The bold numbers marked in the matrix are the exponents $6 = 2^{\bf 1} \cdot 3^{\bf 1} \cdot 5^{\bf 0}, 9 = 2^{\bf 0} \cdot 3^{\bf 2} \cdot 5^{\bf 0}$