Question

Show that if we have an orthogonal set of vectors φ1, . . . , φk, then φ1, . . . , φk are linearly independent as well, i.e.

Answer 1

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.

$\varphi_i\cdot\varphi_j=\begin{cases}\|\varphi_i\|^2&\text{if }i=j\\0&\text{if }i\neq j\end{cases}$

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.