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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 $B$?

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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 $C(A)$.

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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.

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