I have a theoretical question. Why are non-square matrices not invertible?

I am running into a lot of doubts like this in my introductory study of linear algebra.

  • $\begingroup$ A good question, but do you have any thoughts on this? $\endgroup$ Jul 12, 2013 at 0:10
  • $\begingroup$ I'm thinking that it's because of improper linear transformation? The elements don't get mapped properly from an mxn matrix to an nxm matrix $\endgroup$
    – user85362
    Jul 12, 2013 at 0:17
  • $\begingroup$ See my answer. But following with your line of thinking...keep in mind multiplying a function by its inverse must result in the identity matrix. An identity matrix must be square. $\endgroup$ Jul 12, 2013 at 0:20

6 Answers 6


I think the simplest way to look at it is considering the dimensions of the Matrices $A$ and $A^{-1 }$ and apply simple multiplication.

So assume, wlog $A$ is $m \times n $, with $n\neq m$ then $A^{-1 }$ has to be $n\times m$ because thats the only way $AA^{-1 }=I_m$

But it must also be true that $A^{-1 } A=I_m$ but now instead of $I_m$ you get $I_n$ wich is not in accordance with the definition of an Inverse ( see ZettaSuro)

Hence $m$ must be equal to $n$

  • 4
    $\begingroup$ Wow. Just wow.. $\endgroup$
    – Don Larynx
    Nov 25, 2013 at 19:46
  • $\begingroup$ @sigmatau $I_m$ and $I_n$ are both shortened in my book to $I$, the identity matrix. "but now you get $I_n$ which is not in accordance with the definition of an Inverse" what is this definition ? But i guess when the book shortens $I$ for identity matrix, they must have in mind a specific size, either $I_m$ or $I_n$, otherwise $I$ is not well defined. $\endgroup$
    – john
    Jul 31, 2021 at 17:54

Simple answer: because by definition a matrix is commutative with its inverse on multiplication. That is: $A^{-1}$ is a matrix such that $AA^{-1}=I_n$ and $A^{-1}A=I_n$.

For two matrices to commute on multiplication, both must be square.

More complicated answer: There exists a left inverse and a right inverse that is defined for all matrices including non-square matrices. For a matrix of dimension $m\times{n}$, the left and right inverse are defined as follows:

$$A^L:=\{B|BA=I_n\}$$ $$A^R:=\{B|AB=I_m\}$$

If $A^L=A^R$ , by definition $A^L=A^R=A^{-1}$.

  • $\begingroup$ Your $A^L$ and $A^R$ are affine sets. And they are possibly empty. Are they really called the "left inverse" and the "right inverse" And not rather the set of left and right inverses? And does it really answer this question? Also, in your first paragraph, your notation is not correct. You have a matrix on the left. A set on the right. $\endgroup$
    – Julien
    Jul 12, 2013 at 0:23
  • $\begingroup$ @julien Yes, it is worth noting that for any given matrix there is a set of left and right inverses. But I made sure to say a left and right inverse rather than the left and right inverse. $\endgroup$ Jul 12, 2013 at 0:23
  • $\begingroup$ @julien What would be the notation for "B such that ...." without building a set? $\endgroup$ Jul 12, 2013 at 0:26
  • $\begingroup$ Sometimes, a good old sentence in English is much better than symbols, you know. And that's group theory anyway. We say that $A$ is invertible if there exists $B$ such that bla. If it exists, such a $B$ is unique and is denoted by $A^{-1}$. $\endgroup$
    – Julien
    Jul 12, 2013 at 0:28
  • $\begingroup$ @julien True. It's always so tempting to try and write every single thing out with predicate logic and set notation, sometimes I forget we have plain English. $\endgroup$ Jul 12, 2013 at 0:48

Since this question has just been bumped anyway and I feel like I have something else to add, here are my thoughts:

As pointed out by sigmatau we would have $AA^{-1} = I_m$ and $A^{-1}A = I_n$ and he reasons, that we would have $m=n$ then. I think this is an "unnatural" conclusion: Consider the definition of an inverse of a function $f : A\to B$: It is a function $f^{-1} : B\to A$ with $f^{-1}\circ f = id_A$ and $f\circ f^{-1} = id_B$ and we don't require $id_A = id_B$.

So we can and should ask the question, whether there may be an $A^{-1}$ with $AA^{-1} = I_m$ and $A^{-1}A = I_n$, if $n\neq m$. The answer is no:

The matrix $A$ corresponds to a linear map $f : \mathbb{R}^n \to \mathbb{R}^m, x\mapsto Ax$. By the dimension formula we have:

$n = \dim(\ker f) + \dim(\operatorname{im} f)$.

If $A$ has an inverse, then so does $f$ ($y \mapsto A^{-1}y$). Hence $f$ is injective, so $\dim(\ker f) = 0$ and $f$ is surjective, so $\dim(\operatorname{im} f) = m$. But then $n = 0 + m = m$.

  • $\begingroup$ In fact, just use the that $f$ is an isomorphism. Then $m = n$. $\endgroup$ Dec 19, 2017 at 9:30
  • $\begingroup$ Your comments on the inverse of a function notwithstanding (since the domain and codomain of a function are not necessarily identical), sigmatau's argument does work. Note that the inverse of a matrix $A$ is defined to be a matrix $B$ such that $AB = BA = I$. If $A$ is a non square $m\times n$ matrix , then you reach the conclusion after left and right multiplication with $B$ that $I= I_m $ and $I = I_n$ (also note that $B$ is forced to be an $n \times m $ matrix , otherwise $AB \neq BA $). Thus $I_m = I_n$, which is absurd since $m \neq n $. $\endgroup$
    – john
    Oct 29, 2022 at 7:02
  • $\begingroup$ Even more to the point since the inverse $B$ of a matrix $A$ is defined to be commutative, viz., $AB = BA$, this can only occur if $A$ and $B$ are square. So per the definition of inverse, since two non square matrices can't be commutative, a nonsquare matrix cannot be invertible. $\endgroup$
    – john
    Oct 29, 2022 at 7:08

If $A$ and $B$ are $m\times n$ and $n\times k$ matrices respectively, then the rank of $AB$ is less than or equal to both ranks of $A$ and $B$. So, suppose that $A$ is an $m\times n$ invertible matrix, with $m\neq n$. If its inverse is $B$, then $B$ has to be an $n\times m$ matrix, and $AB=I_m$, $BA=I_n$. So, if $n<m$, then the rank of $AB=I_m$ should be $m$, but also less than or equal to the rank of $A$, which is less than or equal to $n$, which is a contradiction. You work similarly in the case $m<n$.

  • $\begingroup$ This argument can be used to prove a more general statement: If a matrix $A$ has right inverse $B$ and left inverse $C$, then $A$ is a square matrix. After that, it can be proved that in such case $B=C$. $\endgroup$
    – Chilote
    Jan 29, 2015 at 5:06

Maybe I am bumping this but I want to add an example for clarity. Other answers state basically that for a matrix to be invertible, their multiplication should be commutative by definition: $\mathbf{A A^{-1} = A^{-1} A}$ and should be equal to identity matrix. This can only be possible with square matrices. But, this proof could be augmented with geometric intuition I guess.

Here is an example that shows how non-square matrices harm the invertibility of a linear transformation.

$$\begin{pmatrix} a & b & -a & & -b \\ a & b & -a & & -b \\ a & b & -a & & -b \end{pmatrix}$$

You can see that this matrix maps from $R^4$ to $R^3$. But as all linear transformations, it maps the zero vector to zero vector. Also see that it maps $(1, 1, 1, 1)$ to zero vector as well. Then, how the inverse transformation should map $(0, 0 ,0)$? Notice that inverse transformation is from $R^3$ to $R^4$. How can we decide whether to map $(0,0,0)$ to $(0,0,0,0)$ or to $(1, 1, 1, 1)$ or to $(2, 2, 2, 2)$ etc. As you can see, this is just a counterexample but I think, someone should prove that non-square matrices can not have inverses because they can not uniquely map. I think proof has got to do with linear dependence and independence.

Here is my more detailed sketch:

Assume an $nxn$ linear transformation $\mathcal{\bar L}$ like

\begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ a_{31} & a_{32} & \cdots & a_{3n} \\ \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{pmatrix}

Now take two arbitrary vectors from $R^n$, say; $x=(x_1, x_2 \cdots , x_n)$ and $u=(u_1, u_2 \cdots , u_n)$. We are going to take a look at the transformation $\mathcal{\bar L} x = u$. After this step, we will manipulate this transformation to get to linearly independent vectors. This part is long and hard to type so I will just give an example for $3x3$ case. I hope you will be able to see how this generalizes.

Here we have a $3x3$ linear transformation $\mathcal{\bar L_3}$ that maps an arbitrary vector $(x, y, z)$ to another vector $(u, v, w)$. Assuming that $\mathcal{\bar L_3}$ is given by:

$$\begin{pmatrix} a & b & c \\ e & f & g \\ k & l & m \end{pmatrix}$$

Then we have the following system of equations:

$$ax + by + cz = u$$

$$ex + fy + gz = v$$

$$kx + ly + mz = w$$

which can be written as

$$ax + by + cz - u = 0$$

$$ex + fy + gz - v = 0$$

$$kx + ly + mz - w = 0$$

which can be written as

$$(a - \frac{u}{3x})x + (b - \frac{u}{3y})y + (c - \frac{u}{3z})z = 0$$

$$(e - \frac{v}{3x})x + (f - \frac{v}{3y})y + (g - \frac{v}{3z})z = 0$$

$$(k - \frac{w}{3x})x + (l - \frac{w}{3y})y + (m - \frac{w}{3z})z = 0$$

We started with our transformation that maps $(x, y, z)$ to $(u, v, w)$ and got ourselves three vectors that are orthogonal to $(x, y, z)$. There can not be four different vectors that are orthogonal to each other in $R^3$. So two of those four vectors must be equal. If you try to set two of them to be equal, you'll see that $(x, y, z)$ is not arbitrary anymore but is determined by your initial transformation and $(u, v, w)$. Since it is uniquely determined, square matrices, as we saw, can have inverses. If you do this, you'll see that there are cases ending up with division by 0. I believe these correspond to case of determinant being 0 but I am not sure and I haven't checked.

You need to generalize this to higher dimensions. Also you need to also show that for non-square matrices, this problem of linearly independent vectors does not arise and we get a linear transformation that maps many vectors to one vector etc. to complete the argument.

I don't count this as a proof because I still think there is something fishy about my argument but it could serve as a way to generate your own counterexamples if you want to convince yourselves.

  • $\begingroup$ Thanks for sharing. Post provided a refreshing perspective on the question. $\endgroup$
    – Dude156
    Sep 14, 2020 at 21:48

Consider the equation AX = B, dimensions for A - m*n, X - n*1 and B - m*1. B can be any vector in space Rm. For AX to always have a solution, A should contain column vectors that make up the complete Rm space (there are n such vectors of m dimensions).

For m > n, We cannot span a complete 3 dimensional space with just 2 vectors of size 3*1.

For m < n, It means we have more variables than the equations itself, it is not possible to find a solution in this case

For m = n, things fall in place perfectly for AX to always have a solution (given column vectors are independent)

Now why inverse of a matrix has anything to do with AX = B, when AX=B always has solution and column vectors in A span the complete Space, X can be presented as X = $A^{-1 }$B. Inverse only comes into discussion when prior conditions are met, and this can only happen when m=n and column vectors of A are independent. Conversely, when taken transpose on both sides the solution has to still hold in that case the row vectors also has to be independent.

Note: For Singular matrices, since the column vectors are dependent, AX=B doesnt have solution for every B. The system becomes inconsistent and X cannot be represented with $A^{-1 }$


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