1

If we multiply a $m \times 0$ matrix with a $0 \times n$ matrix, which $m \times n$ matrix is the result? Is it the $m \times n$ matrix with all zeroes? Or is it some other matrix?

user107952
  • 20,508
  • 4
    This does not really make sense since m x 0 and 0 x n are empty matrices. – Peter Jul 04 '23 at 15:11
  • 1
    The obvious sometimes escapes one. I would go with zeroes. – Somos Jul 04 '23 at 15:14
  • @jjagmath I saw such "empty"-definitions so often (some cases were far stranger than the cases here) that I thought mathemticians would define it this way. Would be good , if this is not the case here , since everything can be exaggerated. – Peter Jul 04 '23 at 15:27
  • If this answer is valid, then for a given field $\mathbb{F}$: (1) for all $m$ there is exactly one $m \times 0$ matrix, which can be viewed as a sequence of $m$ 'rows' of elements of $\mathbb{F},$ each 'row' being a sequence of length $0;$ (2) there is exactly one $0 \times n$ matrix, which can be viewed as a sequence of $n$ 'columns' of elements of $\mathbb{F},$ each 'column' being a sequence of length $0;$ (3) therefore each element of the product is the sum of a sequence of length $0$ in $\mathbb{F},$ which by convention is $0.$ – Calum Gilhooley Jul 04 '23 at 15:53
  • 1
    As noted in this other answer, an $m \times 0$ matrix represents a linear transformation from a $0$-dimensional vector space to an $m$-dimensional vector space. There is only one such linear transformation: the zero map. Matrix multiplication corresponds to composition, and composing the zero linear transformation with any linear transformation will result again in zero. So yes, it's the zero matrix. – John Palmieri Jul 04 '23 at 18:46

1 Answers1

6

Well, by all means it's the zero matrix. Specifically, $$[AB]_{h,k}=\sum_{j=1}^0 A_{h,j}B_{j,k}=0$$

since $\sum_{j=1}^0$ is an empty sum.

This is also consistent with $A$ representing the zero map $\Bbb F^0\to \Bbb F^m$ and $B$ representing the zero map $\Bbb F^n\to\Bbb F^0$.

In order to make sense of matrices with size zero, you need to treat $m\times n$ matrices as a triple $(A,m,n)$, where $A$ is an function in $\Bbb F^{\{1,\cdots,m\}\times\{1,\cdots,n\}}$ and $m,n$ are the two aforementioned natural numbers. The piece of information conveyed by $m$ and $n$ is usually unnecessary, since for $m,n$ positive the function $(m,n)\mapsto\{1,\cdots,m\}\times\{1,\cdots,n\}$ is injective, but, if you are to allow $m=0\lor n=0$, then you need to distinguish when the empty matrix is the representation in coordinates of a map $\Bbb F^n\to \Bbb F^0$, $\Bbb F^0\to \Bbb F^m$ or $\Bbb F^0\to\Bbb F^0$.

For the determinant, the zero matrix $0\in\Bbb F^{0\times 0}$ is treated as the identity of $\Bbb F^0$, therefore its determinant is $1$.

Sassatelli Giulio
  • 718