Understanding a Proof Behind "Two Orthogonal Vectors Are Linearly Independent"

Question

In Eric Lengyel's book, Mathematics for 3D Game Programming and Computer Graphics (3^rd Edition), there is a a theorem in the second chapter stating that two orthogonal vectors are linearly independent, along with a proof.

Given two nonzero vectors $e_{1}$ and $e_{2}$, if $e_{1} \cdot e_{2} = 0$, then $e_{1}$ and $e_{2}$ are linearly independent.

The proof is provided as follows.

We suppose that $e_{1}$ and $e_{2}$ are not linearly independent and arrive at a contradiction. If $e_{1}$ and $e_{2}$ are linearly dependent, then there exist scalars $a_{1}$ and $a_{2}$ such that $a_{1}e_{1} + a_{2}e_{2} = 0$. Note that $a_{2}$ cannot be zero since it would require that $a_{1}$ also be zero. Thus, we can write $e_{2} = -(\frac{a_{1}}{a_{2}})e_{1}$. But then $e_{1} \cdot e_{2} = -(\frac{a_{1}}{a_{2}})e_{1}^{2} \neq 0$, a contradiction.

I believe the proof provided by Lengyel, E. is a proof by contradiction. I understand most of the proof, which utilizes the definition of linear independent vectors. I am stumped with the last sentence of the proof. How did the proof jump to $e_{1} \cdot e_{2} = -(\frac{a_{1}}{a_{2}})e_{1}^{2} \neq 0$? And why is it considered to be a contradiction?

Answer 1

Since $e_1$ and $e_2$ are orthogonal, $e_1 \cdot e_2 = 0$, and in the other hand, since $e_2 = -\tfrac{a_1}{a_2}e_1$, we obtain that $$e_1 \cdot e_2 = e_1 \cdot \Big( -\frac{a_1}{a_2}e_1 \Big) = -\frac{a_1}{a_2} e_1 \cdot e_1 = -\frac{a_1}{a_2} \|e_1\|^2.$$

Of course, $a_1$ cannot be zero by the same argument of why $a_2$ cannot be zero, so, the number $-\tfrac{a_1}{a_2}$ is non-zero. Also, since $e_1$ is a non-zero vector, it's norm squared is non-zero. Hence $-\frac{a_1}{a_2} \|e_1\|^2$ is a product of non-zero numbers and this is a contradiction: $$0 = e_1 \cdot e_2 = -\frac{a_1}{a_2} \|e_1\|^2 \neq 0.$$

Answer 2

Here it is a direct approach to the problem.

Consider the linear combination of the orthogonal vectors $v_{1}$ and $v_{2}$, where both are different from zero. Thus we have \begin{align*} a_{1}v_{1} + a_{2}v_{2} = 0 & \Longrightarrow \langle a_{1}v_{1}+a_{2}v_{2},v_{1}\rangle = \langle 0,v_{1}\rangle = 0\\\\ & \Longrightarrow a_{1}\langle v_{1},v_{1}\rangle + a_{2}\langle v_{2},v_{1}\rangle = 0 \end{align*}

since $\langle v_{1},v_{2}\rangle = 0$, we conclude that $a_{1}\langle v_{1},v_{1}\rangle = 0$. But $v_{1}\neq 0$ and the inner product is positive definite, which means that $\langle v_{1},v_{1}\rangle \neq 0$. Consequently, $a_{1} = 0$. Similarly, one concludes that $a_{2} = 0$. Thence we conclude that $\{v_{1},v_{2}\}$ is LI indeed.

BONUS

The same reasoning applies to the set $\{v_{1},v_{2},\ldots,v_{n}\}$ of orthogonal vectors.