# $0\times n$ matrices

I think it is desirable to have that $M_{m\times n}\left(\mathbb{K}\right)\not=M_{m'\times n'}\left(\mathbb{K}\right)$ if $m\not=m'$ or $n\not=n'$. In other words, the set of all $m\times n$ matrices on $\mathbb{K}$ should be different from the set of all $m'\times n'$ matrices on the same field when $m\not=m'$ or $n\not=n'$. But if I define $M_{m\times n}\left(\mathbb{K}\right)$ as $\left(\mathbb{K}^n\right)^m$ then $M_{0\times n}\left(\mathbb{K}\right)=M_{0\times n'}\left(\mathbb{K}\right)$ for every $n$ and $n'$. That is, two matrices with zero rows are equal, no matter how many columns they have, because $X^0$ is the set containing the empty tuple, for every set $X$. I could have defined $M_{m\times n}\left(\mathbb{K}\right)$ as $\left(\mathbb{K}^m\right)^n$ instead, but then the problem is with $M_{m\times 0}\left(\mathbb{K}\right)$ and $M_{m'\times 0}\left(\mathbb{K}\right)$.

Two questions:

1. Am I defining matrices correctly?
2. Are $m\times 0$ or $0\times n$ matrices that important?

Thanks.

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2: No, they are not important at all, so why bother? – Hans Lundmark Nov 11 '10 at 18:57

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@Hans: I disagree. Talking about 0xn and nx0 matrices may seem like nonsense but it is a concrete way of talking about an important fact: that the category of vector spaces has a terminal object which is also an initial object. – Qiaochu Yuan Nov 12 '10 at 22:31

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@Qiaochu: Matrices are useful for doing computations with concrete linear transformations. The mappings to and from the trivial vector space are easy enough to think of abstractly, so what is there to gain by trying to represent them by matrices? Anyway, the last paragraph in Arturo's answer sums it up pretty well, I think. – Hans Lundmark Nov 12 '10 at 23:15

## 2 Answers

As Laurent S points out, the sensible way to define $\mathbb{K}^0$ is as the one element set (either $\{0\}$ or $\{\emptyset\}$).

You could likewise argue that since you want $M_{m\times n}(\mathbb{K})$ to correspond to linear transformations from $\mathbb{K}^n$ to $\mathbb{K}^m$, and since $\mathbb{K}^0$ is a perfectly good vector space, then we should define the matrices $M_{0\times n}(\mathbb{K})$ and $M_{m\times 0}(\mathbb{K})$. Then you are certainly forced to define both as the trivial set, since only the trivial linear transformation exists in either case (the zero map for $m=0$, and the trivial inclusion for $n=0$). In that respect, they would fit with the general theory and not require exceptions when describing vector spaces (where you would need to explicitly exclude the zero-dimensional vector space, for instance).

Why, then, are $M_{0\times n}(\mathbb{K})$ and $M_{0\times n'}(\mathbb{K})$ both equal even when $n\neq n'$, if the first "is" a linear transformation from $\mathbb{K}^n$ to $\{0\}$ and the second "is" a linear transformation from $\mathbb{K}^{n'}$ to $\{0\}$, and hence different? The key is the `"`. They are both identified with those maps, but as a vector space itself they are in fact isomorphic, so there is no horrible universe-shattering error that crept in by having the "matrix" rings be equal. Of course, now your theorem that $M_{m\times n}(\mathbb{K})$ is uniquely determined (in some way) by the pair $(m,n)$ is no longer true: you need to say that they are equal if and only if either $(m,n)=(m',n')$ or $mn=m'n'=0$.

Such is life: there is often a special case that needs to be dealt with separately, and your choice is whether you exclude it by design or you exclude it by mention. Here, your choice is between specifying "nonzero" in a lot of theorems about vector spaces, or specifying "or $mn=m'n'=0$" in the theorem about matrices. The latter is usually less disruptive. (In set theory, one often has to deal with the emptyset in a special manner; that corresponds to $0$ here).

As to their interest: they have very little structure to them, so they are not generally interesting in and of themselves. They can show up as part of a general scheme (as above, fitting into the role of representing linear transformations between the standard finite dimensional vector spaces) so that exceptions don't need to be made, but as such they play more the role of "exception-avoiders" than "interesting in and of themselves."

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1. No,I guess the only sensible way of defining $\mathbb{K}^0$ when $\mathbb{K}$ is a field is by $\mathbb{K}^0=0$, since it should be a 0-dimensional vector space.

2. No, I have never seen $m\times 0$-matrices used anywhere, other than here.

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