Inverse Matrix ( If there are two different left inverse for a matrix then the matrix has no right inverse)

Question

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 $n=m$?

Answer 1

• 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.

Answer 2

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

Answer 3

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

$m = n. \tag{13}$