Proof of center of $GL(2,\mathbb R)$

Question

I'm examining a proof that: $Z\bigl(GL(2,\mathbb R)\bigr)= \left\{ \pmatrix{ a & 0 \\ 0 & a } \,\middle|\, a \in \mathbb R\setminus\{0\} \right\} $, where $Z$ denotes the center of the general linear group $GL(2,\mathbb R)$.

I have one question on part of this proof:

Suppose we have a matrix of the form $aI$ where $I$ is the $2 \times 2$ identity matrix, then for any $A \in GL(2,\mathbb R)$:

$(aI)A=aA=Aa=A(aI)$*

Therefore such a matrix lies in $Z(GL(2,\mathbb R))$.

Why at * can we assume $aA=Aa$? Is it because $a$ is the identity matrix and $A$ is invertible so these properties imply multiplicative commutativity?

Sorry if this is a trivial question but help would be great!

Answer 1

Actually, $Aa$ is not a good choice of notation, but if you want to use it, I suggest that you define $Aa$ as being equal to $aA$.

Anyway, observe that both $(a\operatorname{Id})A$ and $A(a\operatorname{Id})$ are equal to $aA$.

Answer 2

This follows from the fact that $a$ is a scalar (not the identity matrix).

Answer 3

I agree with Jose that we should not think about scalar multiplication from the left and scalar multiplication from the right - there is only one scalar multiplication. This scalar multiplication (which I will write on the left) does however have the following very useful property (relation to matrix multiplication):

$$(aA)B = a(AB) = A(aB)$$

For all matrices $A, B$, regardless of whether they commute. Using this and combining it with $IX = XI$ for all $X$ you get your answer:

$$(aI)A = I(aA) = (aA)I = A(aI)$$

where the outer two equalities ar instances of the 'useful property' of scalar multiplication and the inner equality is just $I$ commuting with everything.