Linear algebra, prove matrix inverse proof flawed

[karnten07]

Linear algebra, flawed proof

Homework Statement

Theorem: Every square matrix which has a right inverse is invertible. More precisely: let A \in M_{}nxn (R) and suppose there is a matrix B\in M_{}nxn (R) such that AB = I_{}n; then we have BA=I_{}n 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_{}nxn (R) which have a right inverse in M_{}nxn (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\mapstoG
(a¦b) \mapstoaxb
\foralla,b,c\inG, (a*b)*c = a*(b*c)
\existse \inG \forall a\inG, e*a=a=a*e
a'*a=e
\foralla \inG \exists a'\inG, a*a' = e

For any a\inG There exists precisely one right inverse a' and this is also a left inverse of a. We write a^{}-1 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.

Homework Equations

The Attempt at a Solution

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

 

[karnten07]

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.

 

[Hurkyl]

(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)

 

[karnten07]

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?

 

[karnten07]

Homework Statement

Theorem: Every square matrix which has a right inverse is invertible. More precisely: let A \in M_{}nxn (R) and suppose there is a matrix B\in M_{}nxn (R) such that AB = I_{}n; then we have BA=I_{}n 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_{}nxn (R) which have a right inverse in M_{}nxn (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\mapstoG
(a¦b) \mapstoaxb
\foralla,b,c\inG, (a*b)*c = a*(b*c)
\existse \inG \forall a\inG, e*a=a=a*e
a'*a=e
\foralla \inG \exists a'\inG, a*a' = e

For any a\inG There exists precisely one right inverse a' and this is also a left inverse of a. We write a^{}-1 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.

Homework Equations

The Attempt at a Solution

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

Hi guys, I'm still having real trouble with this question and i have tried to systematically think through it. I have an idea as to why the proof is flawed. Is it because in proposition 1.3b it says:

G, G x G \rightarrow 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.

 

[karnten07]

On second thoughts, i think thr propsotion takes care of this fact because on the next line it says,

(a¦b) \mapstoaxb

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 a-1 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''\inG is another right inverse of a

\Rightarrowa'*(a*a'')=(a'*a)*a'' = a'*e=a' by associativity
=e*a''=a''

I'm really out of ideas, anyone have any?

 

[karnten07]

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?

 

[HallsofIvy]

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

Let G be the set of all matrices in M^nxn(R) which have a right inverse in M^nxn(R). Then G together with matrix multiplication is a group.

You are trying to prove that each such matrix has an inverse but asserting that G is a group is the same as asserting each has an inverse.

 

[karnten07]

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


You are trying to prove that each such matrix has an inverse but asserting that G is a group is the same as asserting each has an inverse.

I see what you are saying, but i thought the main point of the theorem was to show that the right inverse of an element in G is also the left inverse of the element ie. AxA^-1=I and A^-1xA=I

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

 

[karnten07]

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


You are trying to prove that each such matrix has an inverse but asserting that G is a group is the same as asserting each has an inverse.

I see it now, i just had to reread the definition of a group. So it is flawed because the so called proof assumes the theorem to be true in defining the set G and matrix multiplication to be a group. This is because for it to be a group it must meet the condition that:

If x\inG, then y\inG 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!