If $A$ if of full column rank, what about $[A;B]$

Question

I meet a fundamental problem as the following:

Suppose $A$ is full column rank, is $$C = \begin{bmatrix}A \\ B\end{bmatrix}$$ also full column rank? Suppose $B$ has the same number of column as $A$.

I think the answer is yes since $B$ does not influence the independence of columns of $C$ if $A$ has linearly independent columns.

thanks in advance.

Answer 1

You are right, $B$ does not influence the independence of columns of $C$

One way to see that: $C$ is full column rank iff $x\mapsto Cx$ is injective (one-to-one mapping).

To prove that $x\mapsto Cx$ is injective one must show that $Cx=0\Rightarrow x=0$. But we have $\left(\begin{array}{c} 0 \\ 0\end{array}\right)=\left(\begin{array}{c} Ax \\ Bx\end{array}\right)$ with $A$ full column rank (id injective) hence $x=0$, hence $C$ is full column rank.

Answer 2

Say $A$ is a $m \times n$ matrix and $B$ is a $r \times n$ matrix. Given that rank$A=n$, $C$ will also have full column rank because the number of pivots after row reduction of $C$ will still be the same as that of $A$ and since every column of $A$ is a pivot column, so upon row reduction we will have $$C \to \begin{bmatrix}R\\0\end{bmatrix},$$ where $R=\begin{bmatrix}1&0&0 &\dotsb &0\\0&1&0 &\dotsb &0\\\vdots & \ddots & \dotsb\\0&0&0 &\dotsb &1\end{bmatrix}$ is the row reduced echelon form of $A$.