the rank of a matrix that is combined by two

Question

$A$ is a $m\times n$ matrix and $m<n$.

$b$ is a $m\times 1$ column.

$B$ is a $m\times (n+1)$ matrix defined as a combination of $A$ and $b$: $B=(A,b)$

Question: what is the condition for $b$ such that rank($B$)=rank($A$)?

ps. rank($A$) can < $m$.

Many thanks for the helps!

Answer 1

The assumption that $m<n$ is redundant: the condition for the ranks to be equal is that $b$ belongs to the column space of $A$, that is, it is a linear combination of the columns of $A$.

The rank of $(A\ b)$ cannot be less than the rank $r$ of $A$, which is the dimension of the column space of $A$. Choose $r$ columns of $A$ forming a basis; call them $v_1,v_2,\dots,v_r$. If $b$ does not belong to the column space of $A$, then $\{v_1,\dots,v_r,b\}$ is linearly independent and therefore the rank of $(A\ b)$ is $>r$ (actually, exactly $r+1$). If $b$ belongs to the column space of $A$, then $\{v_1,\dots,v_r\}$ is a basis of the column space of $(A\ b)$ as well.

A different way to express this is known as Rouché-Capelli theorem:

The ranks of $A$ and of $(A\ b)$ are equal if and only if the linear system $Ax=b$ has a solution.

The above discussion is essentially a proof of the theorem.