Product of reduced row-echelon matrices is also reduced row-echelon


Show that the product of two reduced row-echelon matrices is also
reduced row-echelon.

That's what I think:

A reduced row-echelon matrix has columns like $e_1 =(1, 0, \cdots , 0)^T$ and $e_2 =(0, 1, 0, \cdots , 0)^T$.
For columns in between $e_n$ and $e_{n+1}$, only the first $n$ entries will be non-zero.

By noticing these two, I can 'imagine' that the product should be reduced row-echelon. But I cannot write down a clear proof for that. Or say, I don't even know how to start my proof. Can someone give me a helping hand?


Exactly $n-1$ nonzero elements if $\det(A)=0$ for every arrangement


Let $x_1,x_2,\dots,x_{n^2}\in\mathbb{R}$ with the property that any $n\times n$ matrix with exactly these elements has determinant $0$. Suppose also that there are at least $n$ distinct elements.
Do these conditions imply that exactly $n-1$ elements are nonzero?

This is equivalent to asking if exactly $n^2-n+1$ of the elements must be zero.
There is some reason to think so - if you have less than $n$ nonzero elements, some row has all $0$s and the determinant is $0$ for every matrix, while if you have exactly $n$ nonzero elements, then the diagonal matrix will always have nonzero determinant. However, perhaps it is possible to have zero determinant for every arrangement if we have more nonzero elements (though this appears unreasonable).
The $n$ distinct elements condition is to avoid just using the pigeonhole principle to force two rows to be the same for every arrangement.