For square matrix, right or left inverse is equivalent to inverse. [duplicate]

Definitions:

• Let $A$ be an $n\times n$ matrix. The $n\times n$ matrix $B(=A^{-1})$ is an inverse for $A$ if $AB=BA=I$.
• Let $A$ be a $k\times n$ matrix.The $n\times k$ matrix $B$ is a right inverse for $A$ if $AB=I$.The $n\times k$ matrix $C$ is a right inverse for $A$ if $CA=I$.
• Here $I$ is identity matrix of appropriate order.

Now,

Theorem 2: For any square matrix $A$, the following statements are equivalent:

1. $A$ has inverse.

2. $A$ has a right inverse.

3. $A$ has a left inverse.

I know that

Theorem 1: If $A$ has a right inverse $B$ and a left inverse $C$, then $A$ is invertible and $B=C=A^{-1}$.

Proof: $C=CI=C(AB)=(CA)B=IB=B.$

But I can't prove Theorem 2 .

marked as duplicate by user1551, user61527, Martin Argerami, Cameron Buie, Hanul JeonNov 9 '13 at 6:03

• When you write that you can't prove Theorem 1, do you mean that you can't prove Theorem 2? If so, can you edit your question accordingly, please? – Gerry Myerson Nov 9 '13 at 4:05
• Oh very sorry . – Silent Nov 9 '13 at 4:16

To see that Theorem 2 holds, note that $(1) \Rightarrow (2), \, (3)$ since, by the given definition of the inverse of $A$, $B$ is such if $BA = AB = I$, showing $B$ is both a left and right inverse of $A$. To see that $(2) \Rightarrow (1)$, note that, since $A$ is $n \times n$, so is $B$ such that $AB = I$; then $\det(B)$ is well-defined, as is $\det(A)$, and we have $\det(A) \det(B) = \det (AB) = I$, whence $\det(A), \, \det (B) \ne 0$. Thus $A^{-1}$ and $B^{-1}$ exist, and by $AB = I$ we have $B = IB = (A^{-1}A)B = A^{-1}(AB) = A^{-1}I = A^{-1}$. Thus $A$ has a two-sided inverse which is in fact $B$. The demonstration that $(3) \Rightarrow (1)$ is similar.