Where does the basis of row space matrix comes from? So what I know: According to Strang's book the row space basis comes from where the pivots are in the upper triangular. We take those rows and therefore we have our basis. My problem lies here: "The row space of A has the same basis with the U because the two row spaces are the same".
Moreover: "It is true that A and U have different rows, but the
combinations of the rows are identical: same space!"
My $$A= \left(\begin{matrix}2 & 4&5 \\ 8& 0& 3\end{matrix} \right)$$ and $$U= \left(\begin{matrix}1 & 0&0.375 \\ 0& 1& 1.0625\end{matrix} \right)$$
When solving a matrix I found a case where my row space of U was different from A but I wanted to be sure. 
So regardless from which matrix I take rows its the same?
e.x: If I say Basis: $$RS= \left(\begin{matrix}1 & 0&0.375 \\ 0& 1& 1.0625\end{matrix} \right)$$ $$=\left(\begin{matrix}2 & 4&5 \\ 8& 0& 3\end{matrix} \right)?$$

"It is true that $A$ and $U$ have different rows, but the combinations of the rows are identical: same space!" with
$$A= \left(\begin{matrix}2 & 4&5 \\ 8& 0& 3\end{matrix} \right)$$ and $$U= \left(\begin{matrix}1 & 0&0.375 \\ 0& 1& 1.0625\end{matrix} \right)$$
Answer: Yes: $(8,0,3) = 8 \cdot (1,0,0.375)$ and 
$(2,4,5)=2\cdot(1,0,0.375)+4\cdot(0,1,1.0625)$; thus, any linear combination of $(8,0,3)$ and $(2,4,5)$ can be written as a linear combination of $(1,0,0.375)$ and $(0,1,1.6025)$.
Similarly, $\displaystyle(1,0,0.375) = {1\over 8}\cdot(8,0,3)$ and 
$$(0,1,1.0625)={1\over4}\cdot (2,4,5) - {1\over16}\cdot (8,0,3). $$
Thus, any linear combination of $(1,0,0.375)$ and $(0,1,1.6025)$ can be written as a linear combination of $(8,0,3)$ and $(2,4,5)$.
