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.

  • $\begingroup$ Plural: matrices. Singular: matrix. $\endgroup$ – vadim123 Jan 25 '18 at 3:57
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    $\begingroup$ The first criteria is used. For more details, see the answer deleted below. $\endgroup$ – астон вілла олоф мэллбэрг Jan 25 '18 at 3:59
  • $\begingroup$ @vadim123 That's a rather vague hint.. could you be more specific? $\endgroup$ – Saad Jan 25 '18 at 7:49
  • $\begingroup$ If you are familiar with the concept of the rank of a matrix then you can tell that $AB$ has rank at most $n$, and hence cannot be bijective as a linear transformation. $\endgroup$ – Jyrki Lahtonen Jan 25 '18 at 16:56
  • $\begingroup$ @Saad, that's not a hint, that's a nitpick about word usage. I don't think it's possible to find a solution that doesn't use rank or dimension or bases. Dave gives a lovely solution, that does depend on dimension. $\endgroup$ – vadim123 Jan 25 '18 at 20:01

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 $v\neq w$.


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.

  • $\begingroup$ Thank you! I will try to prove that it isn't injective without using dimensions. $\endgroup$ – Saad Jan 26 '18 at 4:47

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