[Math] Why is it linearly dependent when the linear combination is zero only with none zero coefficients in 3D

linear algebra

Title says it all. I'm asking the geometrical sense.

I know it is linearly independent if the linear combination of vectors is zero with all the coefficients are zero. And so do dependent.

Independent $\sum a_n\mathrm{v}_n=0$, for $a_n=0$.

Dependent $\sum a_n\mathrm{v}_n=0$, not all $a_n=0$.

For mathematical notation.

I do understand in 2D, but I really don't understand why is this working in 3D.

Best Answer

Those are not clear statements of dependence and independence of the set$\{v_1,\dots v_n\}$.

The set would be dependent if:

There exist $\alpha_i$, not all of which are zero, such that $\sum \alpha_iv_i=0$

The set would be independent if it satisfies the negation, that there will not be such a set of nonzero coefficients:

If $\sum \alpha_iv_i=0$, then all the $\alpha_i=0$

If all the $\alpha_i=0$ then $\sum \alpha_iv_i=0$ holds all the time, so it is not interesting! It is a special case that always works. A linearly independent set is special precisely because you can't get a combination to add up to zero unless you use all zeros (which will always work.)

In any number of dimensions, linear independence expresses the idea that one vector is not in the span of the other vectors.

For example, if $\sum \alpha_iv_i=0$ where at least one of the alphas is nonzero, (say for convenience, $\alpha_1$) then $v_1=\sum_{i=2}^n \alpha_1^{-1}\alpha_iv_i$, and so $v_1$ is generatable by $v_2\dots v_n$. Then we could just throw $v_1$ out, since we know the other $v_i$ can already generate it.

So when a set is linearly independent, it means that each member really does contribute to the vector space they generate. Each element adds something new that can't be produced by the other vectors.

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