Say we have the following variables:
- A, a matrix that is nxn in size containing complex numbers
- B, a matrix that is also nxn in size containing complex numbers
- x, a scalar
If you multiply, does it matter whether you multiply in either of the following orders:
- (AB)x
- (Ax)B
if:
- x is a real number
- x is a complex number
3 Answers
$\begingroup$$x(AB)=(xA)B=A(xB)=A(Bx)=(AB)x=(Ax)B$
When matrices are being multiplied by a scalar element, the order in which multiplication takes place can be disregarded
It runs down to the axiom of commutativity of multiplication in fields (a scalar is element of a field) $$ \forall a, b \in F: a b = b a $$ and the commutative property of scalar multiplication: $$ x A = x(a_1, \dotsc, a_n) = (x a_1, \dots, x a_n) = (a_1 x, \dots, a_n x) = (a_1, \dotsc, a_n) x = A x $$ where $x \in F$ and $A$ is a matrix over $F$. For matrix multiplication we have $$ x \sum_j a_{ij} b_{jk} = \sum_j (x a_{ij}) b_{jk} = \sum_j a_{ij} (x b_{jk}) = \left(\sum_j a_{ij} b_{jk} \right) x $$ or $$ x(AB) = (xA)B = A(xB) = (AB)x $$
$\endgroup$ 3 $\begingroup$Scalar multiplication actually is short hand for multiplication by a diagonal matrix full of the scalar. In other words:
$$sA = \left[\begin{array}{ccc}s&0&0\\0&\ddots&0\\0&0&s\end{array}\right]A$$
You can convince yourself that this will be the same when multiplying from the left as from the right.
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