Here is my answer:
We assume two cases, and work to prove that both assumptions are incorrect, leading to a proof.
- Without loss of generality, assume that $ \mathbf{a} $ is linearly dependent with $ \mathbf{b} $
(similarly that $ \mathbf{a} $ is linearly dependent with $ \mathbf{c} $ and $ \mathbf{c} $ is linearly dependent with $ \mathbf{b} $ ).
- Without loss of generality, assume that $ \mathbf{a} $ is linearly dependent with a linear combination of $ \mathbf{b} $ and $ \mathbf{c} $.
(And similarly, $ \mathbf{c} $ is linearly dependent with a linear combination of $ \mathbf{a} $ and $ \mathbf{b} $ or $ \mathbf{b} $ is linearly dependent with a linear combination of $ \mathbf{a} $ and $ \mathbf{c} $).
Case 1:
Assume $ \mathbf{a} = \alpha\mathbf{b}$ , so $ \mathbf{a} \cdot \mathbf{b}= \alpha\mathbf{b} \cdot \mathbf{b}$
but $\mathbf{a}$ is perpendicular to $\mathbf{b}$, so $ \mathbf{a} \cdot \mathbf{b}= 0 $,
but $ \mathbf{b} \cdot \mathbf{b} = |\mathbf{b}|^2 \neq 0$
Therefore, case 1 is false.
Case 2:
Assume $ \mathbf{a} = \beta\mathbf{b} + \gamma\mathbf{c}$ so,
$ \mathbf{a} \cdot \mathbf{b} = (\beta\mathbf{b} + \gamma\mathbf{c}) \cdot \mathbf{b}$
$ = \beta\mathbf{b} \cdot \mathbf{b} + \gamma\mathbf{c} \cdot \mathbf{b} = \beta(\mathbf{b} \cdot \mathbf{b}) + \gamma(\mathbf{c} \cdot \mathbf{b}) $
$ \mathbf{b} $ is perpendicular to $ \mathbf{c} $ so,
$ \mathbf{a} \cdot \mathbf{b} = \beta(\mathbf{b} \cdot \mathbf{b}) = \beta|\mathbf{b}|^2 \neq 0$
but $\mathbf{a}$ is perpendicular to $\mathbf{b}$, so again this is a contradiction.
Therefore, $\mathbf{a}, \mathbf{b}, \mathbf{c} $ are linearly Independent.
Is this a correct proof?
Many thanks in advance!
Best Answer
Yes, the proof is correct. There is a simpler proof that uses the same idea of taking the dot product of the linear dependence equation and the different vectors, however.
The vectors are independent iff $\alpha a+ \beta b+ \gamma c=0 \implies \alpha,\beta,\gamma=0$
So consider the equation again, $\alpha a+ \beta b+ \gamma c=0$. Now dot both sides with $a$ to get $\alpha=0$. Similarly $\beta=0$ and $\gamma=0$, and so $a,b,c$ are independent.