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.
 A: 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 $E^{-1}$.
A: If you look closely at the right matrix, you'll see that


*

*It conflates the 2-dimensional subspace generated by $\{(1,0,0,0),(0,1,0,0)\}$ into the
1-dimensional subspace generated by $\{(1,0,0)\}$, because it's first two columns are linearly dependent. In particular, it maps vectors of the form $(3x,-x,0,0)$ to the zero vector, i.e. $\ker A \supset \{(3,-1,0,0)\}$. Columns $1,3,4$ are also obviously linearly dependent, and that yields $\ker A \supset \{5,0,6,-1\}$. 

*It's range (i.e. what you call column space I think) is generated by $\{(1,0,0),(0,1,0)\}$, because it's bottom row is all zeros. 
The left matrix is invertible (lower triangular matrix, non-zero diagonal), and it follows that


*

*$\textrm{rng } A = \textrm{span} \{(1,2,5),(0,1,0)\}$. This is simply the image of the range of the right matrix under the left matrix. This is easy to find because the right matrix's image is created by two canonical basis vectors, so we just had to pick the corresponding columns of the left matrix.

*$\textrm{ker } A \supset \textrm{span} \{(3,-1,0,0),(5,0,6,-1)\}$. And because of the dimension theorem, we can conclude from (1) that in fact $\textrm{ker } A = \textrm{span} \{(3,-1,0,0),(5,0,6,-1)\}$.
