Let $\text{col}(\mathbf{A})$ be the range (column space) of matrix $\mathbf{A}$. If $\text{col}(\mathbf{B})\subseteq\text{col}(\mathbf{A})$, then, for matrix $\mathbf{C}=[\mathbf{A}\quad\mathbf{B}]$ choose the correct answer:
a) $\text{rank}(\mathbf{C})=\text{rank}(\mathbf{A})$
b) $\text{rank}(\mathbf{C})=\text{rank}(\mathbf{B})$
c) $\text{rank}(\mathbf{C})$ cannot be specified in terms of $\text{rank}(\mathbf{A})$ and $\text{rank}(\mathbf{B})$
My intuition tells me that a) is the correct one, because if $\text{col}(\mathbf{B})\subseteq\text{col}(\mathbf{A})$, then the columns of $\mathbf{B}$ are spanned by the columns of $\mathbf{A}$, leaving $\mathbf{C}$ with the same number of linearly independent columns as $\mathbf{A}$.
Following that line of thought, if $\text{col}(\mathbf{B})\subseteq\text{col}(\mathbf{A})\Rightarrow\text{rank}(\mathbf{B})\leq\text{rank}(\mathbf{A})$, it seems to imply that $\text{rank}(\mathbf{C})=\max\{\text{rank}(\mathbf{A}),\text{rank}(\mathbf{B})\}$.
Is this correct? Is there a more formal way to solve this question? Is there a general rule for the rank of concatenated matrices? At first I tried showing something with the rank-nullity theorem, but I got confused and dropped it.
Thanks in advance
We indeed have $rank(C)=rank(A)$.
Rank of $C$ is equal to the dimension of the column space of $C$. Columns of $C$ consists of columns of $A$ and columns of $B$. Given that $col(B) \subseteq col(A)$, we have $col(C)=col(A)$. Hence their ranks are equal.
Notice that $rank(B) \le rank(A)$ alone doesn't imply that $rank(C)=\max\{rank(A), rank(B)\}$.
Edit:
For a general $A$ and $B$, and $C=[A, B]$, we do not have $col(C)= col(A) \cup col(B)$. For example let $C=I_2$, then $col(A) \cup col(B)$ are not even a subspace. We can write down a basis that spans the columns of $A$, of which we note that by doing that would also span columns of $B$, which implies that we would have span the column space generated by the columns of $A$ and $B$. Hence $col(C)=col(A)$.
Well, the subspace condition suffices.