About inner product space: prove $\left=0$ 
Let $\beta$ be a basis for a finite dimensional inner product space. Prove that if $\left<x,z\right> = 0\ \forall z \in \beta,$ then $x=0$.

My work: Let $z \in \beta$, given $\left<x,z\right>=0$, since z is a basis, let $x=\{x_1,x_2...x_k\}$, then $\left<x,z\right>=\sum_{i=1}^k x_i z_i=0$. Since $z_i$ forms a basis and is linearly independent, $x_1=x_2=...=x_k=0$. Hence $x=0$.
Can I leave it here?
 A: Another way to look at it: $V = U\oplus U^{\perp}$, where $U$ is a linear subspace of $V$. At your case, $U = V$ and $\textbf{x}\in U^{\perp}$. Thus $\textbf{x}\in\{0\}$, that is to say, $\textbf{x} = 0$.
If you prefer an algebraic approach, let $\dim V = n$ and $\mathcal{B} = \{v_{1},v_{2},\ldots,v_{n}\}$ be a basis for $V$. Thus we have that $x = x_{1}v_{1} + x_{2}v_{2} + \ldots x_{n}v_{n}$. Consequently, one has that
\begin{align*}
\langle x,v_{j}\rangle = \langle x_{1}v_{1} + x_{2}v_{2} + \ldots + x_{n}v_{n},v_{j}\rangle = x_{j}\langle v_{j},v_{j}\rangle = 0
\end{align*}
Since $\{v_{1},v_{2},\ldots,v_{n}\}$ is a basis, it is linear independent. Hence $v_{j}\neq 0$ for all $j\in\{1,2,\ldots,n\}$. Given that the inner product is a positive-definite function, it means that $\langle v_{j},v_{j}\rangle \neq 0$. Finally, we conclude that $x_{j} = 0$ for $j\in\{1,2,\ldots,n\}$. That is to say, $\textbf{x} = 0$.
A: A more straightforward and arguable less informative solution would be:
Since $\beta$ is a basis of $V$, $\langle x,z\rangle=0$ for all $z\in V$. Assume $x$ is nonzero. Then $\langle x,x\rangle=0$ contradicts the positive definiteness of the inner product. 
