# Are one-by-one matrices equivalent to scalars?

I am a programmer, so to me [x] != x—a scalar in some sort of container is not equal to the scalar. However, I just read in a math book that for 1 x 1 matrices, the brackets are often dropped. This strikes me as very sloppy notation if 1 x 1 matrices are not at least functionally equivalent to scalars. As I began to think about the matrix operations I am familiar with, I could not think of any (tho I am weak on matrices) in which a 1 x 1 matrix would not act the same way as a scalar would when the corresponding scalar operations were applied to it. So, is [x] functionally equivalent to x? And can we then say [x] = x? (And are those two different questions, or are entities in mathematics "duck typed" as we would say in the coding world?)

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Could someone please create further tags for this post? I don't have enough reputation yet. (At least "scalars" would be great.) –  Kazark Sep 16 '11 at 4:15
Mathematicians have many, many notions of equivalence. In some of them, the answer is yes. –  Qiaochu Yuan Sep 16 '11 at 4:17
It is indeed sensible to treat $1\times 1$ matrices as scalars (for most applications). The surprise here is that multiplication is actually commutative! –  Ｊ. Ｍ. Sep 16 '11 at 4:18

No. To give a concrete example, you can multiply a 2x2 matrix by a scalar, but you can't multiply a 2x2 matrix by a 1x1 matrix.

It is sloppy notation.

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You could treat the 2x2 matrix as having 1x1 matrix entries, then define multiplication to be the matrix of products of 1x1 matrices! –  The Chaz 2.0 Sep 16 '11 at 4:29
On the other hand, a column vector ($n \times 1$ matrix) can be multiplied by a scalar as well as by a $1 \times 1$ matrix, and then it's convenient not to make any distinction, for example in the middle step of the calculation $(x \cdot n)n = n(n\cdot x) = n(n^t x) = (nn^t)x$. –  Hans Lundmark Sep 16 '11 at 4:39
A $1\times 1$ matrix is useful for identifying a field with $GL(1)$ and thereby embedding it in $GL(n)$... :P –  Tabes Bridges Feb 7 '13 at 18:18