If we have vectors $V$ and $W$ in $\mathbb{R^n}$ and their dot product is $0$, are the two vectors linearly independent?
I can expand $V_1 \cdot V_2 = 0 \Rightarrow v_1w_1+...+v_nw_n = 0$, but I don't understand how this relates to linear independence.
Suppose that $\vec v_1\cdot\vec v_2=0$ for $\vec v_1\neq\mathbf 0$ and $\vec v_2\neq\mathbf 0$. Additionally suppose that $$ \lambda_1 \vec v_{1}+\lambda_2\vec v_2=\mathbf 0\tag{1} $$ Applying $(-)\cdot v_j$ to (1) gives $$ (\lambda_1\vec v_1+\lambda_2\vec v_2)\cdot \vec v_j=(\mathbf 0)\cdot \vec v_j\tag{2} $$ Expanding (2) then gives $$ \lambda_1(\vec v_1\cdot \vec v_j)+\lambda_2(\vec v_2\cdot\vec v_j)=0 $$ which is equivalent to $$ \lambda_j\lVert\vec v_j\rVert^2=0\tag{3} $$ But now $\lVert\vec v_j\rVert\neq0$ since $\vec v_j\neq\mathbf 0$ so dividing (3) by $\lVert\vec v_j\rVert^2$ gives $$ \lambda_j=0 $$ Since the above works for $j=1,2$ we have that $\lambda_1=\lambda_2=0$. Hence $\vec v_1$ and $\vec v_2$ are independent.
Hint: Consider what happens when one of the vectors is zero. On the other hand, if both of them are non-zero, then they cannot be linearly dependent, since the norm of a non-zero vector is non-zero.
Working backwards, $\{u,v\}$ are dependent exactly when $u=kv$ (or the other way around) for some real constant $k$. But then $u\cdot v=(kv)\cdot v=k(v\cdot v)=k\|v\|^2$. This is zero exactly when either $\|v\|=0$, or $k=0$ (and hence $u=0$).
More generally, Yes any orthogonal set of vectors are linearly independent. So if we consider $R^n$ together with the inner product being the dot product then if the set of vectors have dot product of zero then they are linearly indepedent:
Consider the set A = {$x_1$,$x_2$,...,$x_n$}
Now we need to prove that the sets A is linearly indepedent set that is if you take a linear combinations of A then only solution that works is trivial one.
Consider $a_n \in R$
$a_1x_1 + ... + a_nx_n = 0$
do the dot product with $x_1$ notice everything will vanish since we have by assumption dot product is zero and you'll be left with $a_1x_1^{2} = 0$ so $a_1$ = 0 and you can proceed similarly and you'll find out that all constants are zero.