Find bases of matrix without multiplying This question is related to a solved problem in Gilbert Strang's 'Introduction to Linear Algebra'(Chapter 3,Question 3.6A, Page 190).

Q) Find bases and dimensions for all four fundamental subspaces of A if you know that
$A = \begin{bmatrix}1 & 0 & 0\\2 & 1 & 0 \\ 5 & 0 & 1\end{bmatrix}\begin{bmatrix}1 & 3 & 0 & 5\\0 & 0 & 1 & 6\\ 0 & 0 & 0 & 0\end{bmatrix} = LU = E^{-1}R$

Answer given in the text:

This matrix has pivots in columns $1$ and $3$. Its rank is $r=2$.
Column space : Basis $(1,2,5)$ and $(0,1,0)$ from $E^{-1}$

Why does he choose the first two columns of $E^{-1}$ as a basis ? If anything, the pivot columns of $R$ are an obvious choice for basis.

If you think of the product $E^{-1}R$ as the composition of linear applications, then $E^{-1}$ acts "last" and its columns determine somehow the image of $E^{-1}R$ (of course, it also depends on $R$).
More generally, for any functions from any sets that you can compose (assuming $Im(g) \subset D(f)$ where $D(f)$ is the domain where f is defined) then $Im(f \circ g) = f(Im(g)) \subset Im(f)$ because by definition
$Im(f \circ g) = \{ f(g(x)) \mid x \in D(g)\} = \{ f(y) \mid y \in Im(g) \} \subset \{ f(z) \mid z \in D(f) \}$
If $g$ is a constant function, so is $f \circ g$, and even if the image of $f$ can be very large, the image of $f \circ g$ is just a singleton. On the other hand, if $g$ is identity, then $Im(f \circ g) = Im(f)$ is maximal. You can have every cases between.
More specifically here in a linear setting, where $R : \mathbb{R}^4 \rightarrow \mathbb{R^3}$ and $E^{-1} : \mathbb{R^3} \rightarrow \mathbb{R^3}$,
$Im(E^{-1}R)=E^{-1}(ImR)=E^{-1}(span\{Re_1,Re_2,Re_3,Re_4\})$ if $(e_1,e_2,e_3,e_4)$ is the canonical basis of $\mathbb{R^4}$. If one writes $(e'_1,e'_2,e'_3)$ for the canonical basis of $\mathbb{R}^3$, since the last element of each columns is zero in $R$ (in other words : the last line is null), this shows that $Im(R) \subset span\{e'_1,e'_2\}$ and $rank(R) \leq 2$. But the rank of $R$ is at least $2$ (first and last column are free e.g), hence it is exactly $2$, and $Im(R) = span\{e'_1,e'_2\}$. 
Finally, $Im(E^{-1}R)=span\{E^{-1}(e'_1),E^{-1}(e'_2)\}$ which are precisely the two first columns of $E^{-1}$.
