[proofplan] We assume a linear combination $\sum_{i=1}^k c_i v_i = 0$ and show all coefficients must vanish. Taking the inner product of both sides with a fixed $v_j$ and using orthogonality to eliminate all cross terms isolates $c_j \|v_j\|^2$, which forces $c_j = 0$ since $v_j \neq 0$. [/proofplan]

[step:Take the inner product with $v_j$ to isolate $c_j$]Suppose \begin{align*} \sum_{i=1}^k c_i\, v_i &= 0 \end{align*} for some scalars $c_1, \ldots, c_k \in \mathbb{R}$. Fix $j \in \{1, \ldots, k\}$. Taking the inner product of both sides with $v_j$ and using linearity in the first argument: \begin{align*} 0 &= \left\langle \sum_{i=1}^k c_i\, v_i,\, v_j \right\rangle = \sum_{i=1}^k c_i\, \langle v_i, v_j \rangle. \end{align*} Since $v_1, \ldots, v_k$ are pairwise orthogonal, $\langle v_i, v_j \rangle = 0$ for all $i \neq j$. The only surviving term is $i = j$: \begin{align*} 0 &= c_j\, \langle v_j, v_j \rangle = c_j\, \|v_j\|^2. \end{align*} Since $v_j \neq 0$, we have $\|v_j\|^2 > 0$ (positive definiteness of the inner product). Dividing both sides by $\|v_j\|^2$ gives $c_j = 0$. Since $j \in \{1, \ldots, k\}$ was arbitrary, $c_i = 0$ for all $i$. Therefore $v_1, \ldots, v_k$ are linearly independent.[/step]

[guided]To prove linear independence, we must show that the only solution to $\sum_{i=1}^k c_i v_i = 0$ is $c_1 = \cdots = c_k = 0$. The strategy is the same one used in the proof of [Coordinates in an Orthonormal Basis](/theorems/3267): take the inner product with a specific basis vector to "select" one coefficient. Here we take the inner product with $v_j$ to isolate $c_j$. Suppose $\sum_{i=1}^k c_i v_i = 0$ for scalars $c_1, \ldots, c_k \in \mathbb{R}$. Fix any $j \in \{1, \ldots, k\}$. Taking the inner product of both sides with $v_j$: \begin{align*} \langle 0, v_j \rangle &= \left\langle \sum_{i=1}^k c_i\, v_i,\, v_j \right\rangle. \end{align*} The left-hand side is $0$ (since $\langle 0, v_j \rangle = 0$ for any $v_j$). On the right-hand side, linearity of the inner product in the first argument gives: \begin{align*} 0 &= \sum_{i=1}^k c_i\, \langle v_i, v_j \rangle. \end{align*} By pairwise orthogonality, $\langle v_i, v_j \rangle = 0$ whenever $i \neq j$. So every term in the sum vanishes except $i = j$: \begin{align*} 0 &= c_j\, \langle v_j, v_j \rangle = c_j \|v_j\|^2. \end{align*} Now we use the hypothesis that $v_j \neq 0$. The inner product is positive definite: $\langle v_j, v_j \rangle > 0$ whenever $v_j \neq 0$. So $\|v_j\|^2 > 0$, and we can divide both sides by $\|v_j\|^2$ to get $c_j = 0$. Since $j$ was arbitrary, all coefficients are zero. This is exactly the definition of linear independence. What role does the nonzero hypothesis play? If some $v_j = 0$, then $\|v_j\|^2 = 0$ and the equation $c_j \|v_j\|^2 = 0$ holds for any $c_j$ -- we cannot conclude $c_j = 0$. Indeed, the zero vector is orthogonal to everything, so $\{0, v_2, \ldots, v_k\}$ is pairwise orthogonal but not linearly independent (since $1 \cdot 0 + 0 \cdot v_2 + \cdots + 0 \cdot v_k = 0$ is a non-zero linear relation).[/guided]