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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?2010-11-11
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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.2010-11-12
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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.2010-11-12

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