# Are there non-square matrices that are both left and right invertible?

I am aware that invertible square matrices are left invertible and right invertible, and that the left and right inverses are equal. However, I was wondering whether exists a non square $$m\times n$$ matrice $$A$$, so that exist both:

1. An $$n\times m$$ matrix $$B$$ so that $$AB = I_m$$
2. An $$n\times m$$ matrix $$C$$ so that $$CA = I_n$$

I just couldn't think of an example nor of a proof that these two conditions provide that A is necessarily square.

• If $A$ and $B$ are matrices, then each is a matrix, not matrice. Apr 4, 2017 at 19:34
• Note that $\operatorname{rank}(AB) \leq \operatorname{rank}(A) \leq \min\{m,n\}$ and $\operatorname{rank}(CA) \leq \operatorname{rank}(A) \leq \min\{m,n\}$. $AB$ and $CA$ cannot both have full rank. Apr 4, 2017 at 19:36

Suppose you have an $n\times m$ matrix $A$ with $n\neq m$.

If $n<m$ then $rank(BA)\leq rank(A)\leq n$ and so $BA\neq I_m$ for all $B$.

If $m<n$ then $rank(AB)\leq rank(A)\leq m$ and so $AB\neq I_n$ for all $B$.

• Awesome, thanks.
– Noa
Apr 6, 2017 at 13:15
• Great answer! Took me some time to understand it. Hence, summarizing your math in words for others here. Basically rank(A) <= min(n, m), so matrix multiplying B with A will not be able to increase the rank, so it's impossible to get the Identity of max(n, m) x max(n, m) matrix. Feb 5, 2022 at 9:09

This question is a variation on the simpler question of whether $$A \in \mathbb{R}^{m \times n}$$ has a two-sided inverse, in the sense described in this answer, ie. a $$B \in \mathbb{R}^{n \times m}$$ satisfying $$AB = I_{m}$$ and $$BA = I_{n}$$.

Here we are asking if $$A$$ has both a right inverse $$B \in \mathbb{R}^{n \times m}$$ and a left inverse $$C \in \mathbb{R}^{n \times m}$$, without necessarily requiring $$B$$ and $$C$$ to be the same matrix (note $$n \times m$$ is the only possible dimension of any type of inverse of $$A$$ because any identity matrix must be square).

To show that this would require $$m = n$$ we can first use the following result from basic set theory :

Theorem 1

Given $$f : A \rightarrow B$$, if $$\,\exists\;g, h : B \rightarrow A$$ with $$fg = i_{B}$$ and $$hf = i_{A}$$ (ie $$g, h$$ are respectively right/left inverses of $$f$$), then $$f$$ is bijective and $$g = h = f^{-1}$$.

Proof

$$b \in B \Rightarrow f(g(b)) = b \therefore f$$ surjective. $$f(a_{1}) = f(a_{2}) \Rightarrow (hf)(a_{1}) = (hf)(a_{2}) \Rightarrow a_{1} = a_{2} \therefore f$$ injective, $$\therefore f$$ bijective, so $$f^{-1}$$ is well-defined as a function from $$B \rightarrow A$$.  Then $$g = f^{-1}$$, since $$b \in B \Rightarrow (fg)(b) = b \Rightarrow (f^{-1}fg)(b) = f^{-1}(b) \Rightarrow g(b) = f^{-1}(b)$$.  And $$h = f^{-1}$$, since $$b \in B \Rightarrow (hf)(f^{-1}(b)) = f^{-1}(b) \Rightarrow (hff^{-1})(b) = f^{-1}(b)$$, ie $$h(b) = f^{-1}(b)$$. $$\tag*{QED}$$

Secondly we use the following property of linear functions :

Theorem 2

A function $$f : \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}$$ is linear (ie. $$f(x + y) = f(x) + f(y)$$, and $$f(\lambda x) = \lambda f(x)$$) iff it has the form $$f(x) = Ax$$ for some $$A \in \mathbb{R}^{m \times n}$$.

Proof

'$$\Leftarrow$$' follows from standard properties of matrix multiplication. For '$$\Rightarrow$$', choose $$A = [f(e_{1}), \ldots, f(e_{n})]$$ where $$\{e_{i}\}$$ is the standard basis for $$\mathbb{R}^{n}$$. $$\tag*{QED}$$

Then by Theorem (2), if $$f$$ is the linear function $$\mathbb{R}^{n} \rightarrow \mathbb{R}^{m}$$ defined by matrix $$A$$, existence of the right inverse $$B$$ of $$A$$ would imply the linear function $$g : \mathbb{R}^{m} \rightarrow \mathbb{R}^{n}$$ corresponding to $$B$$ satisfies $$fg =$$ identity mapping on $$\mathbb{R}^{m}$$. And existence of the left inverse $$C$$ of $$A$$ would imply the linear function $$h : \mathbb{R}^{m} \rightarrow \mathbb{R}^{n}$$ corresponding to $$C$$ satisfies $$hf =$$ identity mapping on $$\mathbb{R}^{n}$$.

Then the conditions for Theorem (1) apply and thus $$f : \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}$$ is a bijective linear function.

In particular $$f$$ is surjective, so every element of $$\mathbb{R}^{m}$$ has form :

$$f(x_{1}e_{1} + \cdots + x_{n}e_{n}) = x_{1}f(e_{1}) + \cdots + x_{n}f(e_{n})$$

for some $$\{x_{i}\}$$, so that $$\{f(e_{i})\}$$ is a spanning sequence for $$\mathbb{R}^{m}$$. But from linear algebra (eg. [1, pg 117, 40.7]) this spanning sequence, comprising $$n$$ elements, spans a subspace of dimension $$\leq n$$, and therefore $$m \leq n$$. This same argument applied to the bijective linear function $$f^{-1} : \mathbb{R}^{m} \rightarrow \mathbb{R}^{n}$$ implies $$n \leq m$$ (the inverse of a bijective linear function must be linear - eg. [1, pg 139, 47.3]). Thus we must have $$m = n$$.

References

 Thomas A. Whitelaw (1991), An Introduction To Linear Algebra, 2nd Edition, Blackie.