
#1
Feb1308, 03:41 PM

P: 208

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? 



#2
Feb1308, 03:48 PM

P: 208

Sorry, i can't seem to get latex to work here, it is making things superscript when they should be subscript and the arrows should be maps to.




#3
Feb1308, 04:08 PM

Emeritus
Sci Advisor
PF Gold
P: 16,101

(1) It works better amongst regular text if you use [ itex ] instead of [ tex ].
(2) It works better still if you put an entire expression inside one pair of tags. (instead of putting a single symbol) 



#4
Feb1308, 05:21 PM

P: 208

Linear algebra, prove matrix inverse proof flawed
Is it that proposition 1.3b doesn't explicitly describe a group that is abelian, and for the theorem to be true it requires commutativity of the matrix and it's inverse, ie, that the right and left inverses are the same?




#5
Feb1508, 05:41 PM

P: 208

G, G x G [tex]\rightarrow[/tex] G So applying proposition 1.3b to this case where we consider that all the elements in our G are only square matrices which have right inverses. So when applying matrix multiplication to two elements of our G, does it always produce another square matrix that itself has a right inverse ie. that it is invertible and of the original group G? If not, this might be the flaw?? Thoughts would be very appreciated, thanks. 



#6
Feb1508, 06:11 PM

P: 208

On second thoughts, i think thr propsotion takes care of this fact because on the next line it says,
(a¦b) [tex]\mapsto[/tex]axb i think the ¦ sign might actually be a comma, i probably copied it wrong from the board. So i think this sentence is meant to mean that x assigns an element axb to the ordered pair a, b. There is another part to proposition 1.3b which i thought i didnt need to reproduce because the proof of a1 also being a left inverse of a, had been shown. But here it is anyway because i can't think of anything else that is the flaw: suppose a''[tex]\in[/tex]G is another right inverse of a [tex]\Rightarrow[/tex]a'*(a*a'')=(a'*a)*a'' = a'*e=a' by associativity =e*a''=a'' I'm really out of ideas, anyone have any? 



#7
Feb1508, 10:01 PM

P: 208

Ah, could this be the problem?
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 it says b is a right inverse of c, where c = a'*a but a'*a is not strictly defined to have a right inverse that resides within our G? 



#8
Feb1608, 06:15 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,894

I haven't read through the entire thing but there is an obvious error right at the start:




#9
Feb1608, 05:16 PM

P: 208

Also, it says that each square matrix of the set has a right inverse and so is a group because there is an inverse for each element. I hope i am missing a point there and that you are right because i just want to move on from this question lol, thanks again 



#10
Feb1608, 08:31 PM

P: 208

If x[tex]\in[/tex]G, then y[tex]\in[/tex]G is an inverse element of x if x*y=e and y*x=e where e is an identity element of G. which is what we want to prove so can't make this assumption in doing so. This seems good to me? Thanks HallsofIvy, you guys are clever!! 


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