Is it true that a non-square matrix cannot have both the left and right inverse?

Consider an $m \times n$ matrix $A$. If $n \times m$ matrix $B$ is the right inverse of $A$, viz., $AB = I_m$. Now if given that $A$ also has a left inverse $C$, which is an $n \times m$ matrix such that $CA=I_n$. Then it can be shown that $B=C$ because


So $B=C$ is the inverse of $A$.

Is it true that for all non-square $A$, such an inverse does not exist?

  • $\begingroup$ But couldn't $B$ be $n \times k$ and $C$ be $l \times m$, in which case this argument does not work? $\endgroup$
    – Papayapap
    Commented Apr 1, 2021 at 17:26

2 Answers 2


Yes, it's true (assuming that $n$ and $m$ are finite for this answer). Here's an argument:

  1. to have a right inverse is to have independent columns.
  2. to have a left inverse is to have indep. rows.
  3. the number of indep. cols is the number of indep. rows.


The first column of $AX$ can be thought of as a linear combination of the cols of $A$ by the entries of the first col of $X$; the same goes for each other col. The fact that $AX = I$ then means that the span of the columns of $I$ is included in the span of the columns of $A$, hence, if $A$ is $n \times k$, we know that the cols of $A$ span $\mathbb R^k$; there are $k$ indep. cols.

The same argument applies to rows.

For part 3, one cheap argument relies on determinants: $p$ (column) $n$-vectors are independent if and only if you can select $p$ of the $n$ rows to form a $p \times p$ matrix whose determinant is nonzero. Thus you cannot have more than $n$ independent $n$-vectors. The same applies to row-vectors.

Finally, if the matrix is non-square, the number of independent rows or columns is at most the smaller of the number of rows and number of cols, hence one set or the other is not independent, so either a left or right inverse can't exist.


Another argument: let $K$ be the base field. $A$ is the matrix of a linear map $u$ from $K^n$ to $K^m$. A right inverse $B$ is the matrix of a linear map $v\colon K^m\rightarrow K^n$ such that $uv=\operatorname{id}_{K^m}$, which implies $u$ is surjective.

Similarly, the existence of a left inverse implies $u$ is injective. So if $A$ has both a right and a left inverse, $u$ is an isomorphism, $m=n$ and $A$ is a square matrix.


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