how do i know that if
when A is a matrix
You need to show that.
Matrix Multiplication is not commutative.
Though it is interesting, if something is left inverse is it right inverse, by square matrices? I do not think so, (In general groups the answer is no, but these are matrices, still I think the answer is no).
inverse exists then it is also a right inverse.
I have left it to the reader to show that if one of the inverses exists then
so does the other.
However here is a a nice proof.
That A has a left inverse means that A considered as a function on R^n
(or C^n) whose action is defined by:
then A is one-one and onto from R^n to itself. Then the left inverse B is
a function such that:
But also B takes X to Y such that A(Y)=X, so
Hence a left inverse is a right inverse (or have I made some
assumption that is not valid? )
You use the fact the left and right inverses exist.
But he is asking if one exists then so does the other (which is what you shown right now, though I cannot say I understand your proof. One thing I do understand that you use the fact that A is matrix and not just any general binary algebraic structure thus, it that rule need not always be true in general).
"Matrix Multiplication is not commutative."
Indeed, although for inverses of matrices, it works out. Matrix multiplication is, however, associative.
(AB)^(-1)*(AB) = [(B^(-1)) * (A^(-1)) *(A) * (B)] = B^(-1) * (I) * (B)
= B^(-1) * (I) * (B) = B^(-1) * B = I
Where I is the identity matrix.
A*A^-1 if and only if A^-1*A
(CaptainBlank did. I just wish he was more careful in his proofs like me. So I can understand. He uses the style of Laplace I use the style of Lagrange).
You have shown that,
Which is not necessary by long expansion.
Simply set C=AB
Unless you were trying to show something else which I missed.