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Do anybody know how matrices are built into ZFC theory ? I pretty have no idea how to build them from ZFC axioms. I would like a constructive proof of their existence if possible.

Thanks in advance.

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Lets talk about matrices with real entries. If you are okay with the construction of the reals $\Bbb R$ and the construction of cartesian products $M\times N$ of subsets of natural numbers $M=\{1,...,m\}$ and $N=\{1,...,n\}$, then a $(m\times n)$-matrix $A%$ is nothing more than just a map

$$A\quad:\quad M\times N\to \Bbb R,\quad (i,j)\mapsto A_{ij}$$

which assignes to each pair of row and column number its corresponding entry in the matrix. I guess you are fine with the construction of such maps in ZFC.

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  • $\begingroup$ Great, thanks, just one more thing, can you explain me how the cartesian product is construct under ZFC ? $\endgroup$ – toto Jun 13 '17 at 9:54
  • $\begingroup$ And just one more thing, do operations on matrices are considered as applications or as algorithms ? $\endgroup$ – toto Jun 13 '17 at 9:56
  • $\begingroup$ @toto Essentially for sets $X,Y$ we can define $X\times Y$ as some special subset of $\mathcal P(\mathcal P(X\cup Y)$. But you can probably find this already asked here on Math.SE, e.g. here. $\endgroup$ – M. Winter Jun 13 '17 at 9:57
  • $\begingroup$ @toto As matrices are consideres as maps, matrix operations are consideres as maps that map maps to other maps (oh god, sounds terrible). E.g. the transpose might be something like this $$\cdot^\top\quad:\quad\mathrm{Func}(M\times N, \Bbb R)\to\mathrm{Func}(M\times N, \Bbb R),\quad (\cdot^\top)(A)(i,j)=A_{ji}.$$ I do not know what your distinction is between application and algorithm. $\endgroup$ – M. Winter Jun 13 '17 at 9:59
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    $\begingroup$ @toto ZFC defines only sets and from this maps, but no algorithms. A turing machine (our universal definition of algorithm) is by definition a (partial) map with some special properties. If your next question is how to define turing machines and algorithms in the contex of ZFC, I would recommend asking a new question as this comment section is getting out of hand. I would try to answer it there. $\endgroup$ – M. Winter Jun 13 '17 at 10:13

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