Why are nonsquare matrices not invertible?

Question

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.

Answer 1

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 $n$

Answer 2

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 $A^L=A^R=A^{-1}$.

Answer 3

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 $n = 0 + m = m$.

Answer 4

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 $m<n$.

Answer 5

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.

Answer 6

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 $A^{-1 }$