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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!

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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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  • $\begingroup$ Why should $B$ be assumed to be $n\times m$? Can't it be $\ell\times m$? $\endgroup$ – dejongbrent Dec 25 '15 at 5:18
  • $\begingroup$ If $B$ is an $\ell \times m$ matrix, then $B \cdot A$ is an $\ell \times n$ matrix. If $B$ is a left inverse of $A$, then $B \cdot A$ is an identity matrix, and thus in particular a square matrix. Therefore $\ell = n$. $\endgroup$ – Jendrik Stelzner Dec 25 '15 at 5:21
  • $\begingroup$ Ah. So a non-square matrix with 1's along the diagonal and 0's everywhere else (a pseudo-identity matrix) isn't acceptable here. $\endgroup$ – dejongbrent Dec 25 '15 at 5:23
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I think this is true. Let $f:V\rightarrow W$ be a linear transformation with a left inverse, i.e., it's injective. If $f$ is not surjective then there are always multiple ways to define a left inverse.

To be more precise, let $\{a_i: i\in I\}$ be a basis of $V$. Then $\{f(a_i): i\in I\}$ is linearly independent. We may extend $\{f(a_i): i\in I\}$ to a basis $\{f(a_i),b_j : i\in I,j\in J\}$. Then every map $g:W\rightarrow V$ satisfying $g(f(a_i))=a_i$ (for all $i\in I$) is a left inverse of $f$. If $f$ is not surjective then $f$ cannot be unique because $J$ is nonempty and we can map the $b_j$'s to whatever we like.

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In a ring of matrices, yes it does. If a matrix $A$, or any element of a ring, has a unique left inverse, say $BA=I$, then $$(I-AB+B)A=A-(AB)A+BA=A-A(BA)+BA=0+I=I.$$ The uniqueness implies that $I-AB+B=B$, so $AB=I$, and $B$ is also the unique right inverse. Obviously $A$ has to be square, otherwise $AB$ is meaningless.

It is interesting to notice that if $A$ is a square matrix with entries in a field, say $A\in M_n(K)$, then $B$ is a left inverse for $A$ if, and only if, it is a right inverse, in which case it is the only two-sided inverse.

In fact, $M_n(K)$ is a finite dimensional vector space over $K$ and one can consider the linear map $\varphi: M_n(K)\to M_n(K)$ given by $\varphi(X)=AX$. Since $BA=I$, this map is injective, hence bijective thanks to the rank-nullity theorem. In particular, there is a unique $C \in M_n(K)$ such that $\varphi(C)=I$. Clearly, $C=B$ is the only two-sided inverse for $A$.

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