In previous linear algebra classes, the definition we used for a basis requires linear independence between the vectors, but that is provably equivalent to having n vectors whose span is R^n.
pkukreja
Here, (B), (D) and (E) are not basis because:
(B): we need 2 vectors in the span of R2
(D): we can't express vectors orthogonal to the given vectors
(E): just 2 vectors form a basis for R2, not 3.
(A) and (C) are a basis.
PlanteurJMTLG
@Cake It is equivalent. If for instance $e_{n-1}$ and $e_n$ are not linearly independent, $span(e_1, \dots, e_{n-1}, e_n) = span(e_1, \dots, e_{n-1})$, therefore $\dim(span(e_1, \dots, e_{n-1}, e_n)) \leq n-1 < \dim(\mathbb{R}^n)$.
keenan
I agree with @PlanteurJMTLG! (Though find their handle pretty hard to type! :-))
In previous linear algebra classes, the definition we used for a basis requires linear independence between the vectors, but that is provably equivalent to having n vectors whose span is R^n.
Here, (B), (D) and (E) are not basis because: (B): we need 2 vectors in the span of R2 (D): we can't express vectors orthogonal to the given vectors (E): just 2 vectors form a basis for R2, not 3.
(A) and (C) are a basis.
@Cake It is equivalent. If for instance $e_{n-1}$ and $e_n$ are not linearly independent, $span(e_1, \dots, e_{n-1}, e_n) = span(e_1, \dots, e_{n-1})$, therefore $\dim(span(e_1, \dots, e_{n-1}, e_n)) \leq n-1 < \dim(\mathbb{R}^n)$.
I agree with @PlanteurJMTLG! (Though find their handle pretty hard to type! :-))