Flaw in proof

• Sep 18th 2011, 09:15 AM
page929
Flaw in proof
I need to find the flaw in the following proof. Can anyone help me?

Suppose A has a right inverse (say B)
AB = I
A^(T)AB = A^(T)
B = (A^(T)A)^(-1) (A^(T))
BA = (A^(T)A)^(-1) (A^(T)A) = I
Therefore B is also a left inverse of A
• Sep 18th 2011, 10:26 AM
Opalg
Re: Flaw in proof
Quote:

Originally Posted by page929
I need to find the flaw in the following proof. Can anyone help me?

Suppose A has a right inverse (say B)
AB = I
A^(T)AB = A^(T)
B = (A^(T)A)^(-1) (A^(T))
BA = (A^(T)A)^(-1) (A^(T)A) = I
Therefore B is also a left inverse of A

That argument assumes that \$\displaystyle A^{\textsc t}\!A\$ is invertible. In the case where A and B are matrices, you can check (by considering determinants) that \$\displaystyle A \$ and \$\displaystyle A^{\textsc t}\$ (and hence their product) are indeed invertible. But in some other situations, for example if A and B are linear operators on an infinite-dimensional space, the result fails, and you can have operators which have a right inverse but not a left inverse.