abstract algebra - If a matrix has a unique left inverse then does it necessarily have a unique right inverse (which is the same inverse)?

I know if a matrix has a left and right inverse then the inverses are the same and are (is) unique and the original matrix is a square matrix, thus if I have a matrix which has multiple left inverses for example then it has no right inverse and is a non-square matrix. But if a matrix has a unique left inverse then does it necessarily have a unique right inverse? So basically does it have to be a square matrix to have unique inverse from one side? (I'm guessing yes since an underdetermined linear equation system has either no or infinite solutions but I need confirmation.) Thank You all in advance!

Answer

It has to be a square matrix. This can be shown by using basic properties about linear equation systems:

Let $A$ be a $m \times n$-matrix. An $n \times m$-matrix $B$ is a left inverse of $A$ if and only if
$$\sum_{k=1}^m B_{ik} A_{kj} = (B \cdot A)_{ij} = I_{ij} = \delta_{ij} \quad \text{for every $1 \leq i,j \leq n$}.$$
This results in a linear equation system in the variables $B_{ij}$. It has $mn$ variables and $n^2$ equations, so for the uniqueness of the solution we must have $n^2 \geq nm$ and thus $n \geq m$.

On the other hand we have $BAx = x$ for every $x \in K^n$, so the linear equation system $By = x$ has a solution for every $x \in K^n$. By looking at the echelon form of $B$ we see that this can only happen if $n \leq m$ (if $n > m$ then the echelon form of $B$ contains zero rows). So we must have $n = m$, so $A$ and $B$ are both square matrices.