<s>[INST] Do the spaces spanned by the columns of the given matrices coincide? Reviewing linear algebra here. 

Let $$A = \begin{pmatrix} 
1 & 1 & 0 & 0 \\ 
1 & 1 & 0 & 0 \\ 
0 & 0 & 1 & 0 \\ 
0 & 0 & 1 & 1 
\end{pmatrix} \qquad \text{and} \qquad
B = \begin{pmatrix} 
1 & 0 & 0 \\ 
1 & 0 & 0 \\ 
0 & 1 & 0 \\ 
0 & 0 & 1 
\end{pmatrix}.$$ 
Is the space spanned by the columns of $A$ the same as the space spanned by the columns of $B$? 

My answer: Yes. For $A$, we have 
$$C(A) = 
\left\{\left.a_1\begin{pmatrix} 
1 \\ 
1 \\ 
0 \\ 
0 
\end{pmatrix}+ a_2 
\begin{pmatrix} 
1 \\ 
1 \\ 
0 \\ 
0 
\end{pmatrix}+ a_3 
\begin{pmatrix} 
0 \\ 
0 \\ 
1 \\ 
1 
\end{pmatrix}+ a_4 
\begin{pmatrix} 
0 \\ 
0 \\ 
0 \\ 
1 
\end{pmatrix} = 
\begin{pmatrix} 
a_1 + a_2 \\ 
a_1 + a_2 \\ 
a_3 \\ 
a_3 + a_4 
\end{pmatrix} \,
\right\vert a_i \in \mathbf{R}\right\}\text{.} 
$$ 
Denote $a_1 + a_2 = c_1$, $a_3 = c_2$, and $a_3 + a_4 = c_3$. Then since $a_4$ is arbitrary, any vector in $C(A)$ can be described as $(c_1, c_1, c_2, c_3)^{\prime}$. 
For $B$, we have 
$$ 
C(B) = 
\left\{\left.b_1\begin{pmatrix} 
1 \\ 
1 \\ 
0 \\ 
0 
\end{pmatrix}+ b_2 
\begin{pmatrix} 
0 \\ 
0 \\ 
1 \\ 
0 
\end{pmatrix}+ b_3 
\begin{pmatrix} 
0 \\ 
0 \\ 
0 \\ 
1 
\end{pmatrix} = 
\begin{pmatrix} 
b_1 \\ 
b_1 \\ 
b_2 \\ 
b_3 
\end{pmatrix} \,
\right\vert b_i \in \mathbf{R}\right\}\text{.} 
$$ 
Thus, any vector in $C(B)$ can be described as $(b_1, b_1, b_2, b_3)^{\prime}$, the same form as a vector in $C(A)$.
 [/INST] There is a very elementary solution. By applying the column transformations  $C_1\rightarrow C_1-C_2$ and $C_3 \rightarrow C_3-C_4$ to $A$ you get $$A^*=
\begin{bmatrix}
   0 & 1 & 0 & 0 \\
    0 & 1 & 0 & 0  \\
    0 & 0 & 1 & 0 \\
    0 & 0 & 0 & 1
   \end{bmatrix}$$
If you look at the colums of $A^*$ and $B$ they are the same. (except of the zero vector which does not contribute anything to the span). And $A^*$ and $A$ are column equivalent. So they have the same span. 
</s>