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karnten07
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Linear algebra, flawed proof
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.
Does this have something to do with matrix multiplication being associative and distributive but not always commutative?
Homework Statement
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.
Homework Equations
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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