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For matrix multiplication to work, you have to multiply an $m \times n$ matrix by an $n \times p$ matrix, so we have $$\bigg(m \times n\bigg)\bigg( n\times p \bigg).$$

But what about a $1 \times 1$ matrix? Is this just a scalar? But every matrix can be mulitplied by a scalar; so do $1 \times 1$ matrices break the rule?

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Actually you can multiply an $m \times n$ matrix by an $n \times p$ matrix to produce an $m \times p$ matrix. – Henry May 9 '12 at 16:50
Here's a situation in which it is useful to view a scalar as the trace of a $1\times1$ matrix:… – Michael Hardy May 9 '12 at 16:56
@Henry ah of course, minor error, I'll edit that in – user26069 May 9 '12 at 17:11
up vote 36 down vote accepted

A $1\times 1$ matrix is not the same thing as a scalar. There is a one to one correspondence between them, but $[a]$ and $a$ are two different things, the first being a matrix and the second one being a scalar. The confusion comes from the fact that the rings $R$ and $M_1(R)$ are isomorphic, and so $M_n(R)$ is isomorphic to $M_n(M_1(R))$. When you talk about treating $[a]$ as a scalar, you are implicitly using this isomorphism.

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You might consider up-voting the question. – Michael Hardy May 9 '12 at 16:53

Matrix multiplication can be regarded as a generalization of the vector product.

An $n$-dimensional row vector can be multiplied by an $n$-dimensional column vector. The result is a scalar value known as the dot product (a special case of the inner product).

These two vectors can also be regarded as matrices: an $1\times n$ multiplied by $n\times 1$. The result should then be a $1\times 1$ matrix, and the element of that matrix will be that dot product.

There is certainly room for regarding $1\times 1$ matrices as scalars, when doing so is convenient.

There is no conflict between the product of a matrix by a scalar, and the product of two $1\times 1$ matrices. For instance $2 \times [3] = [6]$ and also $[2][3] = [6]$.

Of course the ${scalar}\times {matrix}$ case is not restricted by $m\times n/n\times k$ compatibility: it just scales every matrix alement by the scalar. But that semantic difference vanishes in the $1\times 1$ case.

So no rule is really broken! In all cases where you can multiply a matrix by either a scalar or a matrix containing just that scalar, the result is the same.

But scalar multiplication has an additional freedom. Typographically speaking, if you "drop the square brackets" from the $1\times 1$ matrix, you gain the flexibility of doing a different kind of multiplication.

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Yes! The simple answer is that a 1 by 1 matrix is a scalar and a scalar is a one by one matrix. This makes sense because if you regard the dot product of two vectors (which always returns a scalar) as a row vector times a column vector then you always get a 1 by 1 matrix. There is no distinction as they have all the same properties (commutativity, same inverse, associative, etc...), if you want to be "weird", you can leave the [brackets] around the 1 by 1 matrix and say that there is a isomorphism from the 1 by 1 group of matrices to the scalars of F (R or C, etc). Define it as, g([A])=A. If you want to be really neat you need to just check that if the matrix is [0] that it has no inverse, and therefore g([0])=0 and g^-1 does not exist for this particular "matrix".

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Actually, your answer does not address the question, which is whether or not matrix multiplication, where one matrix is of size $1\times 1$, is the same as multiplication by a scalar -- and of course they are not the same... – M Turgeon May 15 '12 at 21:37

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