I have the following question:

Given Matrices $A$ and $B$, the following relation exists between their column spaces:

$$\text{col}(B) \subseteq \text{col}(A)$$

Then, which of the following is true for Matrix $C=[A\,\,\,\,\,B]$?

A) $\text{rank}(C)=\text{rank}(A)$

B) $\text{rank}(C)=\text{rank}(B)$

C) It is not possible to specify $\text{rank}(C)$ in terms of $\text{rank}(A)$ and $\text{rank}(B)$

My guess, and it seems a reasonable one, would be alternative (A), but I don't know how to solve/express it mathematically.

Is my guess correct? Could you walk me through the steps to prove it?

Thanks in advance

  • $\begingroup$ It is true that $\mathrm{rank}(C) = \mathrm{rank}(A)$, since $\mathrm{col}(C)\subset\mathrm{col}(A)$ and vice versa. $\endgroup$ – Math1000 Jan 4 at 22:19
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    $\begingroup$ What is the column space of C? $\endgroup$ – Doug M Jan 4 at 22:20
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    $\begingroup$ $C \binom{x}{y} = Ax + By$. Hence ${\cal R}C = {\cal R}A + {\cal R} B$. You are given ${\cal R}B \subset {\cal R} A$. $\endgroup$ – copper.hat Jan 4 at 22:31
  • $\begingroup$ Is it correct that if $\text{col}(B) \subseteq \text{col}(A)$, then $\text{rank}(B) \leq \text{rank}(A)$ and the rank of a concatenated matrix is equal to the maximum rank among the concatenating matrices? Is this a way to solve it? $\endgroup$ – bertozzijr Jan 4 at 23:11

Yes, that's correct.

The columns of $C$ are just the columns of $A$ followed by the columns of $B$, which are all included in $\mathrm{col}(A)$, hence $$\mathrm{col}(C) \ =\ \mathrm{col}(A)$$ so their dimensions - the ranks - coincide.


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