Inverse Matrix ( If there are two different left inverse for a matrix then the matrix has no right inverse) My question is very short: 
Let $A$ $n$ by $m$ matrix. 


*

*If there are two different left inverse for A then the matrix has no right inverse, right?

*If $A$ is invertible, then must  $A$ be $n=m$?

 A: *

*Right. If $A$ has a left inverse, it is injective. If it had a right inverso too, then it would be surjective and therefore a linear isomorphism. But a linear isomorphism only has one left inverse.

*Right, once again. Just apply the rank-nullity theorem, for instance.

A: an inverse matrix can only be a square matrix $Mat_{n \times n}$.
If the matrix $A$ has a left inverse matrix, then that means that the matrix $A$  also has a right inverse = left inverse = $A'$.
every invertable matrix has one only one inverse matrix. 
suppose that $B$ is the inverse matrix of $A$, then: $A B = B  A = I$, as $I$ is the Identity matrix
A: Suppose we work over an arbitrary base field $\Bbb F$, with
$A \in M_{n \times m}(\Bbb F); \tag 1$
then $A$ may be regarded as a linear map
$A: \Bbb F^m \to \Bbb F^n; \tag 2$
if 
$B_1, B_2: \Bbb F^n \to \Bbb F^m \tag 3$
are left inverses for $A$, then
$B_1A = B_2A = I_m, \tag 4$
where
$I_m: \Bbb F^m \to \Bbb F^m \tag 5$
is the identity mapping on $\Bbb F^m$; from (4),
$(B_1 - B_2)A = B_1A - B_2A = 0_{m \times m}; \tag 6$
now if
$C \in M_{m \times n}(\Bbb F) \tag 7$
satisfies
$AC = I_n, \tag 8$
then from (6) and (8),
$B_1 - B_2 = (B_1 - B_2)I_n  = (B_1 - B_2)AC = ((B_1 - B_2)A)C = 0_{m \times n}, \tag 9$
whence
$B_1 = B_2, \tag{10}$
therefore, if
$B_1 \ne B_2, \tag{11}$
there is no such $C$.
If $A$ is invertible, there is a single $B$ with
$AB = BA = I_m = I_n, \tag{12}$
so tacit in this assumption is the fact we must have 
$m = n. \tag{13}$
