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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 $B$?

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).

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  • $\begingroup$ Is there a way of characterizing all matrices whose column space contains a given subspace? Is this preserved by similarity? $\endgroup$
    – MSIS
    Feb 27 at 14:30
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    $\begingroup$ If you mean by your second question, is column space preserved by similarity, the answer is no. For example, $\pmatrix{0&1\\0&0}$ is similar to its transpose, but they have different column spaces. $\endgroup$ Feb 27 at 18:37
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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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