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According to Wikipedia-

In mathematics, a field is a set on which addition, subtraction, multiplication, and division are defined, and behave as when they are applied to rational and real numbers.

Now, when we work with matrices we define addition,subtraction,multiplication operations (not division) on matrices.

So,going by the above definition, does a matrix qualify to be a field as we define the mentioned operations on them?

I have two doubts-

  • The definition mentions "...a field is a set..." . I know a set is any well defined collection of numbers. A matrix is also a collection of numbers but not well defined. Is this a reason why we cannot call a matrix a set?
  • Division of matrices is not defined. Is this also a reason that eliminates the possibilities of a matrix to be called a set?

Thanks for any help!

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  1. A matrix is definitely not a set of its entries, but all matrices of a specified size over a specified field is a set. This is what we call "the set of matrices".
  2. Indeed, this set does not form a field, by two reasons:
    1. Matrix multiplication is not commutative: in general, $AB \neq BA$.
    2. In a field, all non-zero elements are invertible. This is not the case for matrices: a matrix $A$ is invertible only if $\det A$ is invertible (is not zero, if the scalars form a field), and there do exist non-zero matrices with zero determinant.

So, the structure that the set of matrices forms is a non-commutative ring.

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  • $\begingroup$ Does invertible mean division in $\mathbb R$? $\endgroup$ – tatan Mar 21 '18 at 14:47
  • $\begingroup$ @tatan Invertible means $\exists \frac{1}{x}$ in the ring of scalars. For example, in $\mathbb{R}$ all non-zero elements are invertible. However, in $\mathbb Z$ only $\pm 1$ are invertible, and a matrix with integer coefficients is invertible (with the inverse also having integer coefficients) iff its determinant is $\pm 1$. $\endgroup$ – lisyarus Mar 21 '18 at 15:17
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As you mentioned, the set of all matrices is not a field.

There are many problems with the matrices which keep it far from being a field.

Even if we consider the set of all invertible $n\times n$ matrices along with the $n\times n$ matrix of $0$ we still do not have a field.

The big problem is that even if $AB$ and $BA$ are defined, we do not have $AB=BA$ for all $A$ and $B$.

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