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

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 $m<n$.