Let \(\{\varphi_i~|~i\in\mathbb N,1\le i\le k\}\) be a set of orthogonal vectors. By definition of orthogonality, any pairwise dot product between distinct vectors must be zero, i.e.
\(\varphi_i\cdot\varphi_j=\begin{cases}\|\varphi_i\|^2&\text{if }i=j\\0&\text{if }i
eq j\end{cases}\)
Suppose there is some linear combination of the \(\varphi_i\) such that it's equivalent to the zero vector. In other words, assume they are linearly dependent and that there exist \(c_i\in\mathbb R\) (not all zero) such that
\(\displaystyle\sum_{i=1}^kc_i\varphi_i=c_1\varphi_1+\cdots+c_k\varphi_k=\mathbf 0\)
(This is our hypothesis)
Take the dot product of both sides with any vector from the set:
\(v_j\cdot\displaystyle\sum_{i=1}^kc_i\varphi_i=c_1\varphi_j\cdot\varphi_1+\cdots+c_k\varphi_j\varphi_k=\varphi_j\cdot\mathbf 0\)
By orthogonality of the vectors, this reduces to
\(c_j\|\varphi_j\|^2=0\)
Since none of the \(\varphi_i\) are zero vectors (presumably), this means \(c_j=0\). This is true for all \(j\), which means only \(c_i=0\) will allow such a linear combination to be equivalent to the zero vector, which contradicts the hypothesis and hence the set of vectors must be linearly independent.
