Problem: Take an arbitrary orthonormal basis in $\Bbb R^n$, say $\{ x_1,...,x_n \}$. Suppose that a vector $v \in \Bbb R^n$ has the property that $v \bullet x_i=0$ for all $1 \leq i \leq j$. Then I want to show that $v = 0$ (algebraically).
This seems like an easy proposition using the standard basis but I am not sure how to proceed with an arbitrary basis or arbitrary orthonormal basis. Any hints appreciated.
We can write $v=\sum_{i=1}^na_ix_i$ for some $a_1,\ldots,a_n\in\mathbb R$. Since the basis $\{ x_1,...,x_n \}$ is orhtonormal, for each $j$ we have $$ 0=v \bullet x_j= \left(\sum_{i=1}^na_ix_i\right)\bullet x_j=a_j(x_j\bullet x_j)=a_j.$$ Thus $a_j=0$ for all $j$, and therefore $v=0$.
This is a consequence of the useful fact that if $\{x_1,\dots,x_n\}$ is an orthonormal basis for $\mathbb{R}^n$ then $$ v=(v\cdot x_1)x_1+\dots+(v\cdot x_n)x_n$$ for all $v\in\mathbb{R}^n$.
To prove this, note that since $\{x_1,\dots,x_n\}$ is a basis we can write $v=c_1x_1+\dots+c_nx_n$ for some scalars $c_i$, and then taking the dot product with $x_i$ shows that $c_i=v\cdot x_i$.