Necessary to prove the inverse is Invertible?

Question

I am just starting out on linear algebra and I have come to a section the book that confuses me somewhat.

The authour defines an invertible matrix A as:
"A square matrix A is said to be invertible or non-singular if there is a matrix B of the same size such that
AB=BA=I
Matrix B is called the (multiplicative) inverse of A and is denoted by A^-1 "

The author then goes on to prove the following proposition:
" If A is an invertible matrix then A^-1 is invertible and (A^-1)^-1 = A "

My question is: Is it really necessary to prove this? I think that it should be obvious from the defintion as the inverse A^-1 = B and B's inverse is clearly A since
AB=BA=I is equivalent to BA=AB=I so you could just swich "A" against "B" in the definition?

Or does this proposition really need to be proved to be considered true?

Edit: Before the proof below the author had proved that A^-1 is unique. However, he continues to prove that (A^-1)^-1 = A.

The authors proof that (A^-1)^-1 = A is the following:

" Let B = A^-1 then by the definition of the inverse matrix (above):
A is said to be invertible if there is a B such that AB=BA=I.

We know that B is an invertible matrix because BA=I. Required to prove B^-1=A.
By the definition of the inverse matrix we have BB^-1 = I and BA=I
Equating these gives:
BB^-1 = BA
Left multiplying both sides by B^-1 yields:
B^-1(BB^-1) = B^-1(BA)
(B^-1B)B^-1 = (B^-1B)A
I B^-1 = I A
B^-1 = A
Hence B^-1 = A which means (A^-1)^-1 = A because B=A^-1. Thus we have our result "

As you can see it is kind of long, at least I think so...

Answer 1

I understand what you are saying, but I think one problem is this: how do you know the inverse of a matrix is unique? It might not make sense to write $A^{-1}$ because this could represent multiple matrices. The fact that the inverse of a matrix is unique requires proof, at the very least. Here it is below:

Suppose $A$ is an $n \times n$ matrix and there exist $n \times n$ matrices $B$, $C$ such that $AB = BA = I$ and $AC = CA = I$. We want to show $B$ must equal $C$, and then that would mean the inverse of a matrix is unique (i.e., a matrix can't have two inverse matrices).

Well, $B = BI = B(AC) = (BA)C = IC = C$, and this shows that $B = C$. Note that I used the assumptions that $AC = I$ and $BA = I$, and also that matrix multiplication is associative (i.e., B(AC)=(AB)C).

So, now it makes sense to call the inverse of $A$ as $A^{-1}$, because we know if a matrix is invertible, by the above proof there is only one inverse.

Finally, as you said, we know if $A$ is invertible, then there is a unique matrix $A^{-1}$ such that $AA^{-1} = A^{-1}A = I$. But this is precisely the statement we need to say $A^{-1}$ is invertible and its inverse is $A$. So, you are right. Once we know the inverse is always unique, the same statement that says $A$ is invertible also shows $A^{-1}$ is invertible and its inverse is $A$.

Answer 2

Unfortunately it is necessary, and in this particular situation it is obvious (as you said) that it is the case. After all, that is usually taught in introduction courses.

The point that this rather simple proof hides in plain sight is the need for commutativity with the inverse (AB = BA = I) for this result to hold.

In non-asssociative algebras it could be the case that $ab = I$ but $ba \neq I$ ($b$ is called right identity of $a$ in this case.)

Loops are an interesting case of non-associative structures which we are not sure when an element $b$ is invertible if $ab= I.$

It is an interesting question to wonder why it is important to prove that, and it is equally interesting to understand why.

Hope this helps.