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

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

• What would this representation accomplish? What do you want it for? – Qiaochu Yuan May 1 '14 at 2:46
• Well it goes a bit deeper and this might sound a bit crack-pot but starting off to develop a foundation showing how numbers are entangled, and what a measurement on the "entanglement" collapses to. Indeed it gets used for something much less insane than the above, but since I need points in a PID/UFD? - so as to uniquely represent objects I am looking in to associating the natural numbers to these points. – Relative0 May 1 '14 at 2:54
• While I myself cannot understand why OP want to "turn natural numbers into matrices", the idea of replacing multiplication of natural numbers (or an element of an algebra) by tensor product is actually one of the key ingredient in modern representation theory, or rather categorification. In the case of of natural number, consider the category of vector spaces, then isomorphism classes of objects can be identified with (decategorify) natural numbers (being the dimension of space), while the operations: direct sum and tensor product, decategorify to addition and multiplication. – Aaron May 1 '14 at 12:23
• Let $I_n$ be the $n\times n$ identity matrix. Then $I_n\otimes I_m=I_{nm}$. This seems rather meaningless and random. What's the point? – blue May 1 '14 at 19:50
• What's wrong with considering the ring $R$ of square matrices with the usual addition and $\otimes$ multiplication? I think sea turtles' embedding $\mathbb Z \to R$ by $1\mapsto I_1$ sounds most reasonable. I think this kind of curiosity is good and should not be deemed useless :) – Peter Patzt May 5 '14 at 10:39

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

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.

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 $6 = 2^{\bf 1} \cdot 3^{\bf 1} \cdot 5^{\bf 0}, 9 = 2^{\bf 0} \cdot 3^{\bf 2} \cdot 5^{\bf 0}$