Is the union of the two given sets linearly independent?

Question

In $\mathbb{R}^D$, we have the following two sets: $A=\{a_1,\dots,a_n\}$ and $B=\{b_1,\dots,b_m\}$ where both $A$ and $B$ are linearly independent such that $\sum_{i=1}^{n}\alpha_i a_i = 0$ implies $\alpha_1=\dots =\alpha_n=0$ and $\sum_{i=1}^{m}\beta_i b_i = 0$ implies $\beta_1=\dots =\beta_m=0$. Moreover, $n+m < D$. Additionally, each $b_i \in B$ is orthogonal to each of $a_1, \dots, a_n$. My aim is to prove or disprove that the union $A \cup B$ is linearly independent. I tried an inductive approach; $\{a_1, \dots, a_n, b_1\}$ is linearly independent. Assuming $\{a_1,\dots,a_n,b_1,\dots, b_k\}$ is linearly independent, I try to show $\{a_1,\dots,a_n,b_1,\dots, b_k,b_{k+1}\}$ is also linearly independent. I do know that $\sum_{i=1}^{n}\alpha_ia_i + \sum_{i=1}^{k}\beta_i b_i =0$ implies $\alpha_1=\dots =\alpha_n = \beta_1 \dots = \beta_k = 0$ due to the induction assumption. Moreover I have that each $b_i$ is orthogonal to each $a_i$ with their dot products giving $0$. But I don't see how to start to prove or disprove the claim at that point. How should one proceed here?

Answer 1

Let $U$ the vector subspace generated by $U$ and $V$ the vector subspace generated by $V$, $A\cup B$ is linearly independent is equivalent to $U\cap V=0$. Suppose that $A\cup B$ is linearly indepent, let $x\in U\cap V, x=u_1a_1+...+u_na_n=v_1b_1+...+v_mb_m$ implies that $xu_1a_1+...+u_na_n=-(v_1b_1+...+v_mb_m)=0$, we deduce that $u_i=v_i=0$.

Suppse that $U\cap V=0$, $u_1a_1+...+u_na_n+v_1b_1+...+v_mb_m=0$ implies that $u_1a_1+...+u_na_n=-(v_1b_1+...+v_mb_m)\in U\cap V$ we deduce that $u_1a_1+...+u_na_n=v_1b_1+...+v_mb_m=0$ and $u_i=v_i=0$.

By using Gram-schmidt you can replace $A$ and $B$, without restricting the generality, and suppose that $a_i$ is orthogonal to $a_j$ for $i\neq j$, and $b_i$ is orthogonal to $b_j$. $x\in U\cap V$ implies that $x=u_1a_1+...+u_na_n=v_1b_1+...+v_mb_m$, this implies that $(x,b_i)=(u_1a_1+..+u_na_n,b_i)=0=(v_1b_1+..+v_mb_m,b_i)=v_i(b_i,b_i)$ implies that $v_i=0$ similarly, $u_i=0$.