Linear Algebra 7a: Three Linearly Independent Vectors in the Plane

[Baffage]

Intuitively, it seems to me that since two non-parallell vectors always span the whole plane they're in, it is impossible to construct a third vector in the same plane linearly independent of the first two, as it would necessarily be within their span. Is this not a proof that having three linearly independent vectors in the plane is impossible?

[michaellewis]

It is not obvious only because it wasn’t specified that we were looking trivial linear combinations. If we apply the restriction that ever coefficient of a linear combination must at least have an absolute magnitude greater than zero, then you can have 3 vectors which are linearly independent. A choice of any two other non parallel vectors would satisfy this.

Now if we don’t have this restriction then trivially, there is no third vector. The maximum number of linearly independent vectors will be the dimension of the R^n.

Here, this restricted span would have a hole at the origin.

[ultramadscientist]

Has anyone tried giving you two orthonormal basis vectors and then one that has components [sqrt(-1), 0]?

[paul]

The linear algebra equivalent of sending us to ask for some tartan paint and a long weight 🤨😁

[akimboanansi]

Initially I thought 3 parallel vectors could result in linear independence but then I remembered in lecture 2f you mentionned that vectors that do not pass through the origin are invalid when doing linear algebra. You mention this birefly however and I was wondering if you could elaborate on why this is? I get the impression that two parallel vectors in the same plane are actually exactly the same vector but I don't really understand why this would be true and the intuition behind parrallel vectors generally.

[clo]

when I play the video in 1.2x speed, this guy sounds almost exactly like Feynman