A question about orthogonal vector sets and linear independence Suppose $V=\{v_1,v_2,...,v_n\}$ and $W=\{w_1,w_2,...,w_m\}$ are linearly independent sets of vectors in $\mathbb{R}^n$ such that $v_i$ is orthogonal to $w_j$ for all $i$ and $j$. 

Prove that their union is also linearly independent.
let X=a1w1+a2w2+...+amwm,Y=b1v1+b2v2+...+bnvn. Because vi is orthogonal to wj for all i and j, so XY=0

Suppose X+Y=0, then X and Y are both zero since XY=0,because X and Y are both linearly independent, so all ai and all bi are zero, hence the union is linearly independent. 

Is my answer correct?
Let $w =\sum_i a_i w_i$, $v=\sum_j b_j v_j$ and suppose $w+v = 0$. Then
$w^T(w+v) = \|w\|^2 = 0$ shows that $w=0$ and hence $v=0$. Since the $w_i, v_j$ are
linearly independent, it follows that $a_i = 0$ and $b_j = 0$.

Note: Since $w_i \bot v_j$ for all $i,j$ we have $w^T v = 0$ Since
$w^T(w+v) = w^T w + w^T v = \|w\|^2$.