Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics I / 1 / a / i

Table of contents
- Solution i

Solution

The vectors are linearly independent if

\sum_{j = 1}^{n} a_{j} v_{j} = 0, a_{j} \in R ⟹ a_{1} = \dots = a_{n} = 0.

They span

V

if every

v \in V

is a linear combination

v = \sum_{j = 1}^{n} a_{j} v_{j}

with real coefficients. They form a basis of

V

if they are linearly independent and span

V

. Equivalently, every

v \in V

has exactly one such representation: existence is the spanning set condition, and subtracting two representations shows that uniqueness is precisely linear independence.

 Ancestors (12)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version