on the verbal explanation of full rank

Question

Consider two full column rank matrices $A$ and $B$. If any non-zero linear combination of $A$ does not lie in the space spanned by $B$, can we say that $[A,B]$ is also of full column rank?

Or, if any linear combination of $A$ is linearly independent of any linear combination of $B$, can we say that $[A,B]$ is of full column rank?

Answer 1

Note: I've submitted an edit to this post to change the hypothesis "If any linear combination of $A$ does not lie in the space spanned by $B$" to "If any non-zero linear combination of $A$ does not lie in the space spanned by $B$".

Denote the column vectors of $A$ by $v_1, \dots, v_m$ and the column vectors of $B$ by $u_1, \dots, u_m$. Then $[A, B]$ has full column rank if the set of vectors $\{v_1, \dots, v_m, u_1, \dots, u_n\}$ is linearly independent. Consider the linear combination

$$a_1v_1 + \dots + a_mv_m + a_{m + 1}u_1 + \dots + a_{m + n}u_n$$

where $a_1, \dots, a_{n + m}$ are scalars and not all $a_i$ are zero. If $a_{m + 1} = a_{m + 2} = \dots = a_{m + n} = 0$, then this linear combination is non-zero since $A$ has full column rank. Similarly, if $a_1 = a_2 = \dots = a_m = 0$, this linear combination is non-zero since $B$ has full column rank. So, suppose there are some $a_i, a_j$, $1 \leq i \leq m$, $m + 1 \leq j \leq m +n$ such that $a_i, a_j \neq 0$. Then if our linear combination above is zero, we have

$$a_1v_1 + \dots + a_mv_m = -(a_{m + 1}u_1 + \dots + a_{m + n}u_n)$$

where $a_1v_1 + \dots + a_mv_m \neq 0$ since $A$ has full column rank. But then we have a non-zero vector in the column space of $A$ contained in the column space of $B$, a contradiction. Therefore, our set of vectors $\{v_1, \dots, v_m, u_1, \dots, u_n\}$ is linearly independent and $[A, B]$ has full column rank.

Answer 2

Based on your explanation, I considered in terms of matrix.

Let $A=[a_1,...a_p]$ and $B=[b_1,...,b_q]$ and both $A$ and $B$ are of full column rank.

Let us denote the linear combination of $A$ and $B$ given by $x=A \alpha+B \beta$ where $\alpha$ is $p \times 1$ and $\beta$ is $q \times 1$. When $[A,B]$ is full rank, $x=0$ only when $\alpha=\beta=0$.

Since $B$ is of full column rank, we can define $M_B=I-B(B'B)^{-1}B'$. Multiplying this to $x$, we have $M_B x=M_B A\alpha=0$. Since $A$ does not lie in the column space of $B$, it follows that $M_BA \neq 0$. Hence, $M_B x=M_B A\alpha=0$ only when $\alpha=0$.

Is my proof correct or missing something?