span

the span of $\text{[math]}$ is the set of linear combinations $\text{[math]}$ . it is denoted $\text{[math]}$ .
[cite:;taken from @calc_hubbard_2015 chapter 2.4.3]

the span of a given set of vectors is the collection of all the possible linear combinations of the vectors of the set, i.e.:
let $\text{[math]}$ be a vector space, let $\text{[math]}$ and $\text{[math]}$ , then $\text{[math]}$

let $\text{[math]}$ be vectors in $\text{[math]}$ ; let $\text{[math]}$ be the $\text{[math]}$ matrix $\text{[math]}$ . then

$\text{[math]}$ are linearly independent iff the row-reduced matrix $\text{[math]}$ has a pivotal $\text{[math]}$ in every column.
$\text{[math]}$ span $\text{[math]}$ iff $\text{[math]}$ has a pivotal 1 in every row.

[cite:;taken from @calc_hubbard_2015 theorem 2.4.5]

let $\text{[math]}$ , then $\text{[math]}$

let $\text{[math]}$
$\text{[math]}$ we have proven that $\text{[math]}$ is a subset of $\text{[math]}$ , now with that out of the way we need to check the 3 conditions that need to be satisfied for a subset to be a subspace
condition 1: $\text{[math]}$
$\text{[math]}$ condition 2: addition closure
let $\text{[math]}$ so there exist $\text{[math]}$ so that:
$\text{[math]}$ condition 3: multiplication closure
let $\text{[math]}$ so there exist $\text{[math]}$ , such that $\text{[math]}$
$\text{[math]}$

by definition of span, a span of a set of vectors is the collection of all the possible linear combinations using said vectors
the span of a single vector would just be its own line expanded across the entire 3d region, because we cant reach other dimensions by constant multiplication or by addition of the vector to itself
$\text{[math]}$

  from sage.plot.plot3d.shapes2 import Line
  my_plot = Line([(0,0,0),(1,1,1)], thickness=2, arrow_head=True, color='red') +\
      Line([(-2,-2,-2),(2,2,2)], thickness=1) +\
      text3d("vector", (0.6,0.6,0.6), rotation=45.0).rotate((0,0,0.5),0.5) +\
      text3d("vector space", (-0.5,-0.5,-0.5), rotation=45.0).rotate((0,0,0.5),0.5)
  <<sage3dhandler>>

the span of 2 linearly independant vectors is however more interesting because it would be visualized as a grid across the 3d space
for simplicity, we take the relatively simple vectors $\text{[math]}$ that lie on the $\text{[math]}$ grid which covers the $\text{[math]}$ and $\text{[math]}$ axes
$\text{[math]}$ $\text{[math]}$

given the vectors
$\text{[math]}$ is $\text{[math]}$ in the span of the other three? we now reduce
$\text{[math]}$ since $\text{[math]}$ contains a row of 0's, span.html says that $\text{[math]}$ do not span $\text{[math]}$ . however, $\text{[math]}$ is still in the span of those vectors: $\text{[math]}$ contains no pivotal 1, so Theorem 2.2.1 says there is a solution to $\text{[math]}$ . but the solution is not unique: $\text{[math]}$ has a column with no pivotal 1, so there are infinitely many ways to express $\text{[math]}$ as a linear combination of $\text{[math]}$ . for example,
$\text{[math]}$ is $\text{[math]}$ in the span of $\text{[math]}$ , $\text{[math]}$ , $\text{[math]}$ ?
is $\text{[math]}$ in the span of $\text{[math]}$ , $\text{[math]}$ , $\text{[math]}$ ?
[cite:;taken from @calc_hubbard_2015 example 2.4.6]

are the vectors
$\text{[math]}$ linearly independent? do they span $\text{[math]}$ ? the matrix
$\text{[math]}$ so by span.html, part 1, the vectors are linearly independent. by part 2, these three vectors also span $\text{[math]}$ .
vectors that are not linearly independent are linearly dependent.
[cite:;taken from @calc_hubbard_2015 example 2.4.7]