Do scalars commute across matrices?

Question

$A,B,C$ are matrices that work together, lets just assume they are all $n\times n$, and $a$ is a scalar.

E.g. does $aABC=AaBC=ABaC=ABCa$, I imagine this is the case, but I wanted to verify, and maybe a quick reason why would be good.

Can't really show working since I am asking for a property, so please don't downvote for that...

Yes, basically $(c \mathbf{A}){ij} = c \left(\mathbf{A}\right){ij}$ for some constant $c$. — Autolatry, Mar 31 '15 at 11:18
Another way to express this, is that the matrices $cI$ commute with every $n \times n$ matrix. It turns out these are the ONLY matrices that commute with every $n \times n$ matrix, so feel lucky. — David Wheeler, Mar 31 '15 at 16:41

score 5 · Answer 1 · answered Mar 31 '15 at 19:19

Yes, it is true that scalar multiplication commutes with matrix multiplication. In order to see this, you could try and make the computation coordinate-wise.

Say the coordinates of your matrices are given by $A=(a_{ij})_{n\times n}$ and $B=(b_{ij})_{n\times n}$, then as you should probably know, the multiplication is given coordinate-wise by $$ AB=(\sum_{k=1}^n a_{ik}b_{kj})_{n\times n}. $$

If you insert a given scalar $c\in\mathbb{R}$, then everything boils down to the usual commutativity of the real numbers. Indeed,

$$ cAB=(\sum_{k=1}^n ca_{ik}b_{kj})_{n\times n}=(\sum_{k=1}^n a_{ik}cb_{kj})_{n\times n}=AcB. $$

Naturally, this extends to $n$-fold multiplications.

Do scalars commute across matrices?

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