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.
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