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 $C(A)$.

Solutions Collecting From Web of "Do the spaces spanned by the columns of the given matrices coincide?"

You are correct. One can also verify this by checking that $A, B$ are column-equivalent, or equivalently, that is, that $A^T, B^T$ are row-equivalent. One can check the latter algorithmically by checking that the reduced row echelon form of the $A^T, B^T$ coincide (at least after padding with an appropriate number of zero rows to the smaller matrix).

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.