It's easy to prove that if a $n \times n$ matrix $A$ has a unit LU decomposition $A=LU$, then the span of the rows of $A$ is the same as the span as the rows of $U$. Numerical experimentation seems to imply that this can be generalised: the first $k$ rows of $U$ seem to span the same as the first $k$ rows of $A$ in every example I've tried. However I can't seem to prove this. Every proof I've tried seems to break down as the product $LU$ involves working with the columns of $U$ rather than the rows, and I can't seem to rectify this.
Linear algebra is a weak point of mine, so any hint or partial solution would be much appreciated.
Let $e_i$ be the $i$-th standard unit vector.
Let $a_i^T$ be the $i$-th row of the matrix $A$, that is $a_i^T=e_i^TA$
Let $l_i^T$ be the $i$-th row of the matrix $L$ and let $u_i^T$ be the $i$-th row of the matrix $U$.
$$a_i^T=e_i^TA=e_i^TLU=l_i^TU=\sum_{j=1}^n l_{ij}u_j^T=\sum_{j=1}^i l_{ij}u_j^T$$
where in the last equality, I made use of the property of the lower triangular matrix of $L$.