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Is there such a thing?

I suppose there are two cases in which they may exist: over a finite field or infinite matrices (but let's stick to countable matrices then).

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  • $\begingroup$ Zero matrices of any dimension over any field. You can construct a matrix with one element per column in a countably infinite dimensional vector space over a countable field. Why are you interested in this? $\endgroup$
    – Zack Wolske
    Commented May 25, 2012 at 1:36

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The parity check matrix of a Hamming code, for example $$\pmatrix{1&0&0&1&1&0&1\cr0&1&0&1&0&1&1\cr0&0&1&0&1&1&1\cr}$$ almost has this property; all it's missing is an all-zero column. (This is over the field of 2 elements.)

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Obviously, up to column-permutation, there is one such matrix for each possible range. One simply takes all the elements of the range and makes those the columns. Thus, you are asking about subspaces of vector spaces over finite fields, with an ordering of the elements. Subspaces of vector spaces are obviously very well understood. Ordering the elements doesn't seem to add anything to that story.

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