If the dot product between two vectors is $0$, are the two linearly independent?

Solution 1:

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.

Solution 2:

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.

Solution 3:

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 $u=0$).