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1. The problem statement, all variables and given/known data

Theorem: Every square matrix which has a right inverse is invertible. More precisely: let A [tex]\in[/tex] M[tex]_{}nxn[/tex] (R) and suppose there is a matrix B[tex]\in[/tex] M[tex]_{}nxn[/tex] (R) such that AB = I[tex]_{}n[/tex]; then we have BA=I[tex]_{}n[/tex] as well.

The object of this exercise is to explain why the following proof is flawed:

Proof: Let G be the set of all matrices in M[tex]_{}nxn[/tex] (R) which have a right inverse in M[tex]_{}nxn[/tex] (R). Then G together with matrix multiplication is a group. Now proposition 1.3(b) implies the theorem:

Proposition 1.3b:

Let G be a group

G, GxG[tex]\mapsto[/tex]G

(a¦b) [tex]\mapsto[/tex]axb

[tex]\forall[/tex]a,b,c[tex]\in[/tex]G, (a*b)*c = a*(b*c)

[tex]\exists[/tex]e [tex]\in[/tex]G [tex]\forall[/tex] a[tex]\in[/tex]G, e*a=a=a*e

a'*a=e

[tex]\forall[/tex]a [tex]\in[/tex]G [tex]\exists[/tex] a'[tex]\in[/tex]G, a*a' = e

For any a[tex]\in[/tex]G There exists precisely one right inverse a' and this is also a left inverse of a. We write a[tex]^{}-1[/tex] for the inverse of a.

Proof of proposition 1.3b:

Let a' be a right inverse of a

(a'*a)*(a'*a)=a'*(a*(a'*a)) by associativity

=a'*((a*a')*a)

=a'*(e*a) because a' is a right inverse of a

=a'*e because e is an identity element

Let b be a right inverse of c:=a'*a

c*b

=(c*c)*b

=c*(c*b) by associativity

=c*e since b is a right inverse of c

=c because e is an identity element

Hence a' is a left inverse of a

Note: proposition 1.3b is what is given in the lecture notes.

2. Relevant equations

3. The attempt at a solution

Does this have something to do with matrix multiplication being associative and distributive but not always commutative?

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# Linear algebra, prove matrix inverse proof flawed

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