[Math] Let a, b, c be three nonzero vectors, any two of which are perpendicular. Prove that these 3 vectors are linearly independent.

linear algebra

Here is my answer:

We assume two cases, and work to prove that both assumptions are incorrect, leading to a proof.

  1. 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} $ ).

  1. 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.

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