I've got the following excercise in my textbook:
Prove that if $ A$ is an $m \times n$ matrix, $B$ is an $n \times m$ matrix and $ n < m$, then $AB$ is not invertible.
to which I need to construct a proof only using the concepts of row-equivalence and matrix multiplication (no vector theory, rank etc).
My Proof #1:
Let $Bx=0$ be a system of $n$ linear equations in $m$ unknowns.
We know that
- if $n<m$, then the system has non-trivial solution. Therefore, the system $ABx=0$ also has non-trivial solution (i.e a $p$-parameter family of solutions, non unique).
- if $AB$ is row-equivalent to $I$, then the system $ABx=0$ has only the trivial solution (i.e $x=0$) and vice-versa.
Since $ABx=0$ is shown to have non-trivial solution, $AB$ is not row equivalent to $I$. Hence $AB$ does not have an inverse.
My proof #2:
Consider the matrix $A$. Since $m>n$, the row reduced echelon form of $A$ would have atleast one row of zeroes. Therefore $AB$ would also have atleast one row of zeroes which isn't row equivalent to the identity matrix and hence not invertible.
Is this good?
