# Rank of $\mathbf{C}=[\mathbf{A}\quad\mathbf{B}]$ given $\text{col}(\mathbf{B})\subseteq\text{col}(\mathbf{A})$

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

• @RodrigodeAzevedo fixed! thx – bertozzijr Dec 21 '19 at 14:15

## 1 Answer

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.

• 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 $\text{rank}(\mathbf{C})=\max\{\text{rank}(\mathbf{A}),\text{rank}(\mathbf{B})\}$? – bertozzijr Dec 21 '19 at 13:53
• I edited my answer. – Siong Thye Goh Dec 21 '19 at 14:05