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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
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  • $\begingroup$ Scalar multiplication is commutative, so no, it doesn't make a difference. $\endgroup$
    – Doug M
    May 27, 2016 at 20:14
  • $\begingroup$ @DougM I knew it was commutative, but I didn't know whether it was associative with matrices and whether it plays a role that $x$ is in $\mathbb{C}$ and A and B are in $\mathbb{C^{n \cdot n}}$ $\endgroup$
    – coolcat007
    May 27, 2016 at 23:33
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    $\begingroup$ @M.Vinay Thanks for pointing that out. I indeed know that in a lot of cases the brackets are only needed when multiple characters need to be \somefunctionhere'd. I just do it out of force of habit :P $\endgroup$
    – coolcat007
    May 28, 2016 at 2:43

3 Answers 3

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$x(AB)=(xA)B=A(xB)=A(Bx)=(AB)x=(Ax)B$
When matrices are being multiplied by a scalar element, the order in which multiplication takes place can be disregarded

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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 $$ x(AB) = (xA)B = A(xB) = (AB)x $$

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  • $\begingroup$ Ok, that is a very sound explanation for real numbers. I verified that it is also true for complex numbers: a+ib = (a b, -b a) -> (a+ib)(c+id) = (a b, -b a)(c d, -d c) = (ac-bd ad+bc, -(ad+bc) ac-bd) = ac-bd + i(ad+bc). (c+id)(a+ib) = (c d, -d c)(a b, -b a) = (ca-db cb+da, -(cb+da) ca-db) = (ac-bd ad+bc, -(ad+bc) ac-bd) = ac-bd + i(ad+bc) QED $\endgroup$
    – coolcat007
    May 27, 2016 at 23:09
  • $\begingroup$ I also overlooked that you mentioned $∀a,b∈F:ab=ba$, Also, I forgot the formatting for the matrices, can you fix that? I don't have the right to edit after 5 min. $\endgroup$
    – coolcat007
    May 27, 2016 at 23:13
  • $\begingroup$ Complex numbers form a field, so they have commutative multiplication and more. $\endgroup$
    – mvw
    May 27, 2016 at 23:19
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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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