linear algebra - Let $A,B$ be $m times n$ and $n times m$ matrices, respectively. Prove that if $m > n$, $AB$ is not invertible

We haven't done anything about rank or dimensions or linear dependence / basis or determinants. Possible related facts :

1. A matrix is invertible iff it is bijective as a linear transformation.




2. An invertible matrix is row-equivalent to the identity matrix.




3. A matrix has a right inverse iff it has a left inverse.

Also, invertability is only defined for square matrices.

Answer

Since $A$ is an $m\times n$ matrix and $B$ is an $n\times m$ matrix, the product $AB$ is an $m\times m$ matrix. Suppose $m>n$ then the operator associated to left multiplication by $B$ is not injective because there are more columns than rows, so the kernel is always nontrivial (i.e. there are more column vectors than there are entries in those column vectors, so they must be linearly dependent). Said another way: the linear operator $T:\Bbb R^m\to\Bbb R^n$ with $T(v)=Bv$ is not injective because $m>n$, so the domain has higher dimension than the codomain. So there exists vectors $v,w\in\Bbb R^m$ such that $Bv=Bw$ but $v\neq w$.

Spoiler:

Thus, $$Bv=Bw\implies A(Bv)=A(Bw)\implies (AB)v=(AB)w$$ but $v\neq w$. Hence, the operator associated with left multiplication by $AB$ is not injective.