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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

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  • $\begingroup$ @RodrigodeAzevedo fixed! thx $\endgroup$ – bertozzijr Dec 21 '19 at 14:15
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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.

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  • $\begingroup$ so basically you mean that I could have solved it by starting from $\text{col}(\mathbf{C})=\text{col}(\mathbf{A})\cup\text{col}(\mathbf{B})$, and, as $\text{col}(\mathbf{B})\subseteq\text{col}(\mathbf{A})$, $\text{col}(\mathbf{A})\cup\text{col}(\mathbf{B})=\text{col}(\mathbf{A})$? What else needs to be satisfied in order to have $\text{rank}(\mathbf{C})=\max\{\text{rank}(\mathbf{A}),\text{rank}(\mathbf{B})\}$? $\endgroup$ – bertozzijr Dec 21 '19 at 13:53
  • $\begingroup$ I edited my answer. $\endgroup$ – Siong Thye Goh Dec 21 '19 at 14:05

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