I know that the inverse matrix of a square matrix exists iff its determinant isn't 0.
What about a non-square matrix?
Is there any theorem about the existence of a left- or right-inverse matrix of a non-square matrix?
It seems there does not exist a determinant of a non-square matrix as Determinant of a non-square matrix
If $A$ is a $m \times n$ and $rg(A) = n$, $A$ has a left inverse matrix $B$ which is a $n \times m$ matrix, thus $BA = I_n$.
If $rg(A) = m$, then $A$ has a right inverse matrix $B$, a $n \times m$ matrix. Thus $AB = I_m$.
If you interpret a $m×n$-matrix $A$ with entries in a ring $R$ as an $R$-linear map $f_A\colon R^n → R^m,~x ↦ Ax$, then $A$ is left-/right-invertible if and only if $f_A$ has the corresonding property.
As it turns out, for $f_A$ being right-invertible is equivalent to being surjective. For $f_A$ being left-invertible, there probably is a nice equivalence, too, but I can’t see it right now. You can at least say that $f_A$ needs to be injective and if you are working over a field, that’s sufficient as well!
Now, $f_A$ is surjective if and only if the columns of $A$ span $R^m$ and $f_A$ is injective if and only if the columns of $A$ are $R$-linearly independent.
https://math.berkeley.edu/~shinms/SP14-54/left-right-invertible-matrices.pdf Let m, n be two positive integers. (1) Suppose m < n (more columns than rows). (i) There are no left invertible m × n matrices. (ii) An m × n matrix is right invertible if and only if its REF has pivots in every row.3 (2) Suppose m > n (more rows than columns). (i) There are no right invertible m × n matrices. (ii) An m × n matrix is left invertible if and only if its REF has pivots in every column.4
Hope this helps
