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…
Best Answer
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 $A$.