Answer to Question #167450 in Linear Algebra for Zeeshan Ali

Question

Prove that if S is an orthogonal set of non-zero vectors. The S is Linearly independent.

Answer 1

Consider the linear combination

$$c_1v_1+c_2v_2+..... + c_kv_k=0$$

Our goal is to show that $c_1=c_2=...=c_k=0$

We compute the dot product of $v_i$  and the above linear combination for each $i=1,2,....,k:$

$$0=v_i.0$$

$= v_i .(c_1v_1+c_2v_2+.....+c_kv_k)$

$=c_1v_i.v_1+c_2v_i.v_2+.....+c_kv_i.v_k$

As S is an orthogonal set, we have $v_i.v_j=0$ if $i \neq j$

Hence all terms but the $i$ -th one are zero, and thus we have

$$0=c_iv_i.v_i=c_i||v_i||^2$$

Since $v_i$  is a nonzero vector, its length $||v_i||$  is nonzero.

It follows that $c_i=0$.

As this computation holds for every $i$ =1,2,…,k

$i$ =1,2,…,k, we conclude that $c_1= c_2 =.....=c_k=0$

Hence the set S is linearly independent.