Proving that orthogonal vectors are linearly independent

Solution 1:

Dot through by a1. We get $$c_1(a1\cdot a1)=0$$ so $c_1=0$. The same holds for the other two constants. (I'm assuming that when you say orthogonal, you are not allowing any vectors to have zero magnitude.)

Solution 2:

Let $(v_1,v_2,v_3)$ be a set of orthogonal non zero vectors. Let $\alpha_1,\alpha_2,\alpha_3\in\Bbb R$ such that $$\sum_{k=1}^3 \alpha_k v_k=0$$ then $$\require{cancel}\alpha_j \cancelto{\ne0}{||v_j||^2}=\bigg\langle \sum_{k=1}^3 \alpha_k v_k,v_j\bigg\rangle=0,\;\;\forall j=1,2,3$$

Conclude.

Remark: You can generalize this result for a set of $n$ orthogonal vectors.

Solution 3:

You are given

$$c_1a_1+c_2a_2+c_3a_3=0$$

As $a_1,a_2,a_3$ are orthogonal, we end up with the following three equalities, where $a\cdot b$ denotes the vector product between vectors $a$ and $b$.

$$(a_1\cdot c_1a_1+c_2a_2+c_3a_3)=c_1(a_1\cdot a_1)=0$$ $$(a_2\cdot c_1a_1+c_2a_2+c_3a_3)=c_2(a_2\cdot a_2)=0$$ $$(a_3\cdot c_1a_1+c_2a_2+c_3a_3)=c_3(a_3\cdot a_3)=0$$

But $(a_k\cdot a_k)>0$, for $k\in\{1,2,3\}$.

We can satisfy the above equalities only if $c_1=c_2=c_3=0$, thus proving that the set of orthogonal vectors are linearly independent.