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If each of the vectors $u, v, w$ in $\Bbb R^3$ is perpendicular to the other two, then the three vectors are linearly independent.

When is this true/ false?

UPDATE: A simple (such that a first year student can understand) solution can be done like this:

By observing that the set is not linearly independent if $u,v,w=0$

If the vectors are orthogonal (pairwise), then: $$u\cdot v=0$$ $$u\cdot w=0$$ and $$v \cdot w=0$$

If $u,v,w$ are linearly independent, then the only solution for: $$c_1u + c_2v + c_3w=0 \ \ \ (1)$$ is $$c_1=c_2=c_3=0$$

Multiply $(1)$ by $u \rightarrow$ $$c_1u^2+c_2(u\cdot v) + c_3(u\cdot w)=0\cdot u$$ and so $c_1=0$ since $u\cdot w=0$ and $u\cdot v=0$ (as they are orthogonal).

We can repeat the same procedure, and we get that the only solution to $(1)$ is trivial, hence they are linearly independent.

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You'll have trouble if at least one is the zero vector. If not, consider $$au+bv+cw = 0,$$apply $\langle \cdot, u\rangle$ to conclude that $a=0$. Then repeat the argument with $\langle \cdot, v\rangle$ and $\langle \cdot, w\rangle$ to get $b=c=0$.

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  • $\begingroup$ Hey just wondering what this notation means? $\langle \cdot, u\rangle$ $\endgroup$ – Hugh Entwistle Aug 22 '16 at 10:43
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    $\begingroup$ @HughEntwistle It means applying function $f(\cdot)=\langle \cdot, u\rangle$ (also written as $f(x)=\langle x, u\rangle$) which is the inner product of $\cdot$ and $u$. Applying it to $au+bv+cw$, you get $f(au+bv+cw)=\langle au+bv+cw, u\rangle = a\langle u,u\rangle$ and if $u$ is not zero vector, then you can conclude that $a=0$. $\endgroup$ – Frenzy Li Aug 22 '16 at 10:52

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