Not so easy question (at least I hope) about matrix inverse

How do we prove that if a matrix has right inverse, then it's the left inverse too. Is it true in all cases?

I've read one beginner level book about algebra and it approaches the problem by defining inverse A' of an invertible matrix A to be such matrix that statisfies both AA'=I and A'A=I. Now this type of definition implies that there may be such matrices A and B such that AB=I but BA<>I. I'm not sure how I can discover two such matrices, so providing example by myself is off.

I've read some abstract algebra too. It seems that proving that all nxn invertible matrices (ones that have right inverse) form a group under matrix multiplication may do the job (we do get cancelation laws), but...

If A has right inverse A' then A' must have it's own right inverse to statisfy the group axioms. How do I prove that such one does exist?

Re: Not so easy question (at least I hope) about matrix inverse

I've looked in my abstract algebra text and just remembered that the group axioms may be stated in alternative forms. So my text says:

1. The set must be closed under the binary operation

2. For all members of the set associativty of the binary operation must hold

3. There must be identity el3ment

4. !! If A is in the group, there must be element A' in the same group such that both (!!) AA'=e and A'A=e hold

ugh....

Re: Not so easy question (at least I hope) about matrix inverse

If A is square and AA' = I , multiply both sides on the right by A

AA'A = IA

Matrix multiplication is associative and IA = A

A(A'A) = A

AX = A

X must be the identity therefore A'A = I

This shows that inverses commute

If A is not square then there is something called 'pseudo inverse'

Re: Not so easy question (at least I hope) about matrix inverse

Another stupid question. How do we know there is no other identity. Element X<>I for which AX=A too.

Re: Not so easy question (at least I hope) about matrix inverse

Suppose A has 2 inverses X and Y then

AX = I

AY = I

using the previous result we know inverses commute

XA = I

multiply both sides on the right by Y

XAY = IY

matrix multiplication is associative and IY = Y

X(AY) = Y

but AY = I

XI = Y

X = Y

This shows the inverse is unique.

There are no stupid questions, don't be too hard on yourself. It's only easy when we know how.

:)

Re: Not so easy question (at least I hope) about matrix inverse

This is actually easy and once I knew the method of proof in relations with the group. It uses subset of group axioms, so it easily applies to our scenario.

I is identity. I.e. AI=IA=A for every square nxn matrix ....(1)

I' is identity too. AI'=I'A=A for every square nxn matrix ...(2)

So In (1) set A=I' to get I'I=I'

and in (2) set A=I to get I'I=I

Re: Not so easy question (at least I hope) about matrix inverse

Oh yes , i just realisd i answered a different question... i thought you were asking why inverses are unique.

Re: Not so easy question (at least I hope) about matrix inverse

And Mulder, for this to work it might be not enough that we have unique identity.

We should be sure that if AX=AY then X=Y (right cancelation if i recall?).

Only then we can cancel A from AX=AI in your proof to get X=I.

Or we must be sure that AX=A has unique solution.

Am I correct?

Re: Not so easy question (at least I hope) about matrix inverse

The above is LEFT cancellation As are on the left. (AX=A).

And if we have LEFT cancellation we have the AA'=I, AA'A=IA=AI, cancel As to get A'A=I. Unfortunatelly, unwilling to accept too much, we cannot assume right out the cancellation works.

And from AA'=I I can only prove directly RIGHT cancellation.

CA=DA , CAA'=DAA', CI=DI, C=D

?/????

probably the way is elementwize proof that if AX=AY then X=Y...