[proofplan]
We prove that an orthogonal set of nonzero vectors is linearly independent by showing that every finite linear combination equalling zero must have all coefficients equal to zero. Given a vanishing linear combination, we take the inner product of both sides with each vector in the set. Orthogonality annihilates all cross terms, isolating each coefficient multiplied by the nonzero squared norm, which forces each coefficient to be zero.
[/proofplan]
[step:Take the inner product of the linear relation with each vector to isolate coefficients]
Let $S = \{v_1, v_2, \ldots, v_m\}$ be a finite orthogonal set of nonzero vectors in $V$, so that $(v_i, v_j)_V = 0$ for all $i \neq j$ and $(v_i, v_i)_V \neq 0$ for each $i$. Suppose
\begin{align*}
c_1 v_1 + c_2 v_2 + \cdots + c_m v_m = 0
\end{align*}
for scalars $c_1, c_2, \ldots, c_m$. Fix an arbitrary index $k \in \{1, \ldots, m\}$. Taking the inner product of both sides with $v_k$ and applying linearity in the first argument:
\begin{align*}
0 = \left( \sum_{i=1}^{m} c_i v_i,\, v_k \right)_V = \sum_{i=1}^{m} c_i\, (v_i, v_k)_V.
\end{align*}
Since $S$ is orthogonal, $(v_i, v_k)_V = 0$ for all $i \neq k$, so every term in the sum vanishes except the $i = k$ term:
\begin{align*}
0 = c_k\, (v_k, v_k)_V.
\end{align*}
Since $v_k \neq 0$, we have $(v_k, v_k)_V > 0$ by the positive-definiteness of the inner product. Dividing both sides by $(v_k, v_k)_V$ yields $c_k = 0$.
[guided]
The goal is to show that the only linear combination of the $v_i$ producing the zero vector is the trivial one. How can we extract a single coefficient from a linear combination? The key idea is to use the inner product itself as a "projection tool": taking the inner product with a particular vector $v_k$ and exploiting orthogonality to kill all terms except the one involving $v_k$.
Concretely, suppose $c_1 v_1 + c_2 v_2 + \cdots + c_m v_m = 0$ for scalars $c_1, \ldots, c_m$. Fix any index $k \in \{1, \ldots, m\}$ and take the inner product of both sides with $v_k$. The inner product is linear in the first argument (this is part of the definition of an inner product space), so:
\begin{align*}
0 = \left( \sum_{i=1}^{m} c_i v_i,\, v_k \right)_V = \sum_{i=1}^{m} c_i\, (v_i, v_k)_V.
\end{align*}
Now the orthogonality hypothesis does its work: for every $i \neq k$, we have $(v_i, v_k)_V = 0$, so the entire sum collapses to a single term:
\begin{align*}
0 = c_k\, (v_k, v_k)_V.
\end{align*}
Why can we conclude $c_k = 0$? Because $v_k$ is a nonzero vector, the positive-definiteness axiom of the inner product guarantees $(v_k, v_k)_V > 0$. In particular, $(v_k, v_k)_V \neq 0$, so we may divide both sides by it to obtain $c_k = 0$.
What if some $v_k$ were the zero vector? Then $(v_k, v_k)_V = 0$ and the equation $0 = c_k \cdot 0$ would hold for any $c_k$ -- we could not deduce $c_k = 0$. This is precisely why the hypothesis requires all vectors in $S$ to be nonzero.
[/guided]
[/step]
[step:Conclude linear independence since the index $k$ was arbitrary]
Since $k \in \{1, \ldots, m\}$ was arbitrary, we have $c_k = 0$ for every $k = 1, \ldots, m$. Thus the only linear combination $\sum_{i=1}^{m} c_i v_i = 0$ is $c_1 = c_2 = \cdots = c_m = 0$, which is the definition of linear independence of $\{v_1, \ldots, v_m\}$.
Since every finite subset of an orthogonal set of nonzero vectors is linearly independent, the full set $S$ is linearly independent.
[/step]
