Is my attempt at the following proof correct?
Given that $\beta$ is a basis for a finite dimensional inner product space $V$ . Prove that if $\langle x,z\rangle = 0$ for all $z\in\beta$ then $x = 0$.
Proof. Let $\beta = \{z_1,z_2,...,z_n\}$ and $x = c_1z_1+c_2z_2+\cdot\cdot\cdot+c_nz_n$ then $\langle x,x\rangle = \langle x, c_1z_1+c_2z_2+\cdot\cdot\cdot+c_nz_n\rangle$ and by appealing to the conjugate linearity of $\langle\cdot,\cdot\rangle$ in second component we have $\langle x,x\rangle = \overline{c_1}\langle x,z_1\rangle+\overline{c_2}\langle x,z_2\rangle+\cdot\cdot\cdot+\overline{c_n}\langle x,z_n\rangle$ but from hypothesis $\langle x,z_1\rangle = \langle x,z_2\rangle = \cdot\cdot\cdot = \langle x,z_n\rangle = 0$ consequently $\langle x,x\rangle = 0$ which is equivalent to $x = 0$.
$\blacksquare$
