Linear Independence and Inner Product Spaces

Question

Let $S=\{v_1,v_2,\dots,v_n\}$ be a linearly independent subset of an inner product space $V$, and $w \in V$ where $w$ is orthogonal to each vector in $S$. Prove, using only the definition of linear independence, orthogonal vectors and the inner product space axioms that $S\cup\{w\}$ is also linearly independent.

I'm just not quite sure how to combine the definition of linear independence, orthogonal vectors and the inner product space axioms to show that $S\cup\{w\}$ is also Linearly Independent.

Answer 1

Suppose that $c_1v_1+\dotsb+c_nv_n+c_{n+1}w=0$ for some scalars $c_i$. Take the inner product of the left hand side and right hand side with $w$ and use the fact $w$ is orthogonal to each vector in $S$ to deduce that $c_{n+1}=0$. Finally use the fact that the $\{v_i\}$ form a linearly independent set to deduce that $c_i=0$ for $1\leq i\leq n$.

Answer 2

Suppose

$$w=\sum_{k=1}^n a_kv_k\implies\,\forall\,j\;,\;\;1\le j\le n\;:$$

$$0\stackrel{\text{given}}=\left\langle\,w,v_j\,\right\rangle=\sum_{k=1}^n a_k\langle v_k,v_j\rangle=a_k\langle v_j,v_j\rangle\implies a_k=0$$

since $\;v_1,...,v_n\;$ l.i.

Fill in details, in particular: why is the above enough?