I have the following proposition:
Let $\bf A$ be a square matrix and $\bf R$ be the rref form of A. Then ($\bf R$ has at least one row of zeros $\iff $ $\bf A$ is non-invertible).
My attempt:
Firstly, I will quote the definition of the rref from the same book where the proposition was given:
A matrix is in reduced row echelon form, normally abbreviated to rref, if it satisfies all the following conditions:
If any rows are containing only zero entries then they are located in the bottom part of the matrix.
If a row contains non-zero entries then the first non-zero entry is a 1. This 1 is called a leading 1.
The leading 1’s of two consecutive non-zero rows go strictly from top left to the bottom right of the matrix.
- The only non-zero entry in a column containing a leading 1 is the leading 1.
Note: The reduced row echelon form of $\mathbf A$ is denoted as $\bf R$.
($\Rightarrow $)
We prove by contrapositive, i.e, we need to show that:
If $\bf A$ is invertible, then $\bf R$ doesn't have rows of zeroes.
Suppose $\bf A$ is invertible.
Consider linear system $\bf Ax = O$. Premultiplying by inverse, we have: $$\bf A^{-1}Ax = IO \implies$$ $$\bf x = O \implies$$
It can be seen that lin. system $\bf Ax = O$ has a unique solution, and hence $\bf R$ cannot have a row of zeroes.
($\Leftarrow $)
We prove by contrapositive, i.e, we will show that:
If $\bf R$ doesn't have a row of zeroes, then $\bf A$ is invertible.
If $\bf R$ doesn't have row of zeroes, then $\bf R = I$, and thus:
$$\bf A = E_{1}E_{2}E_{3}.....E_{n-1}E_{n}$$
Where $\bf E_{x}$ is an elementary matrix.
Elementary matrices are invertible, hence their product is invertible as well, and therefore we can conclude that $\bf A$ is invertible.
$\Box$
Is it correct?
Following the advice of the people on the MSE, I'm trying to keep things as concise as possible. Is it "concise" enough? Or I should be a bit more elaborate?
If there are alternatives to prove the proposition, I would be glad to see them!