- #1
Krovski
- 11
- 0
From what I understand, a basis is essentially a subset of a vector space over a given field.
Now what I'm not so sure of is the linearly independence part. If the basis has two linearly independent vectors, then than means they aren't collinear: rather, they wouldn't have the same slope or be generated by each other?
let's say a vector v[itex]\epsilon[/itex]V is (v[itex]_{1}[/itex], v[itex]_{2}[/itex],...,v[itex]_{n}[/itex])
and a vector w[itex]\epsilon[/itex]V is (w[itex]_{1}[/itex], w[itex]_{2}[/itex],...,w[itex]_{n}[/itex])
such that
v [itex]\neq[/itex] cw.
for any constant c[itex]\epsilon[/itex][itex]\textbf{F}[/itex]
These would then be non-collinear which means there are no linear operators that can turn v into w?
Any insight would be greatly appreciated and thank you in advance.
Now what I'm not so sure of is the linearly independence part. If the basis has two linearly independent vectors, then than means they aren't collinear: rather, they wouldn't have the same slope or be generated by each other?
let's say a vector v[itex]\epsilon[/itex]V is (v[itex]_{1}[/itex], v[itex]_{2}[/itex],...,v[itex]_{n}[/itex])
and a vector w[itex]\epsilon[/itex]V is (w[itex]_{1}[/itex], w[itex]_{2}[/itex],...,w[itex]_{n}[/itex])
such that
v [itex]\neq[/itex] cw.
for any constant c[itex]\epsilon[/itex][itex]\textbf{F}[/itex]
These would then be non-collinear which means there are no linear operators that can turn v into w?
Any insight would be greatly appreciated and thank you in advance.