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

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- Sep 18th 2011, 09:15 AMpage929Flaw 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 AMOpalgRe: Flaw in proof
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.