Proving that orthogonal vectors are linearly independent I am trying to prove that $a_1$, $a_2$, $a_3$ are linearly independent.
I am asked to use vector product and prove that if $c_{1}a_{1} + c_{2}a_{2} + c_{3}a_{3} = 0$ then $c_1 = c_2 = c_3 = 0$
I am completely stuck on where to go with this problem. I would think that linearly independent then the null space of the space can only be equal to $0$. Would this prove it?
 A: 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.)
A: 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.
A: 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. 
A: For any orthonormal vector, we know the following 2 properties:

*

*$a_i^Ta_j=0$ when $i\neq j$, because they are perpendicular to each other.

*$a_i^Ta_j=1$ when $i=j$, because the dot product of a vector with itself is the $length^2$ of that vector, and by definition, an orthonormal vector has a length of 1.

If a matrix Q containing only orthonormal column vectors  $a_1  , ..., a_n$, then from the 2 properties we have:$Q^TQ=I$. Just try a few calculations and you will know it's true!
Proofing $c_1=c_2=c_3=0$ is the same as proofing $Qx=0$ only when $x = zero\space vector$.
$Q^TQx=Q^T\times0$
$Ix=0$
$x=zero\space vector$
Thus $c_1=c_2=c_3=0$
