Editing Linear subspace (section)

== Examples ==
=== Example I ===
In the vector space ''V'' = '''R'''<sup>3</sup> (the [[real coordinate space]] over the field '''R''' of [[real number]]s), take ''W'' to be the set of all vectors in ''V'' whose last component is 0.
Then ''W'' is a subspace of ''V''.

''Proof:''
#Given '''u''' and '''v''' in ''W'', then they can be expressed as {{nowrap|1='''u''' = (''u''<sub>1</sub>, ''u''<sub>2</sub>, 0)}} and {{nowrap|1='''v''' = (''v''<sub>1</sub>, ''v''<sub>2</sub>, 0)}}. Then {{nowrap|1='''u''' + '''v''' = (''u''<sub>1</sub>+''v''<sub>1</sub>, ''u''<sub>2</sub>+''v''<sub>2</sub>, 0+0) = (''u''<sub>1</sub>+''v''<sub>1</sub>, ''u''<sub>2</sub>+''v''<sub>2</sub>, 0)}}. Thus, '''u''' + '''v''' is an element of ''W'', too.
#Given '''u''' in ''W'' and a scalar ''c'' in '''R''', if {{nowrap|1='''u''' = (''u''<sub>1</sub>, ''u''<sub>2</sub>, 0)}} again, then {{nowrap|1=''c'''''u''' = (''cu''<sub>1</sub>, ''cu''<sub>2</sub>, ''c''0) = (''cu''<sub>1</sub>, ''cu''<sub>2</sub>,0)}}. Thus, ''c'''''u''' is an element of ''W'' too.

=== Example II ===
Let the field be '''R''' again, but now let the vector space ''V'' be the [[Cartesian plane]] '''R'''<sup>2</sup>.
Take ''W'' to be the set of points (''x'', ''y'') of '''R'''<sup>2</sup> such that ''x'' = ''y''.
Then ''W'' is a subspace of '''R'''<sup>2</sup>.

''Proof:''
#Let {{nowrap|1='''p''' = (''p''<sub>1</sub>, ''p''<sub>2</sub>)}} and {{nowrap|1='''q''' = (''q''<sub>1</sub>, ''q''<sub>2</sub>)}} be elements of ''W'', that is, points in the plane such that ''p''<sub>1</sub> = ''p''<sub>2</sub> and ''q''<sub>1</sub> = ''q''<sub>2</sub>. Then {{nowrap|1='''p''' + '''q''' = (''p''<sub>1</sub>+''q''<sub>1</sub>, ''p''<sub>2</sub>+''q''<sub>2</sub>)}}; since ''p''<sub>1</sub> = ''p''<sub>2</sub> and ''q''<sub>1</sub> = ''q''<sub>2</sub>, then ''p''<sub>1</sub> + ''q''<sub>1</sub> = ''p''<sub>2</sub> + ''q''<sub>2</sub>, so '''p''' + '''q''' is an element of ''W''.
#Let '''p''' = (''p''<sub>1</sub>, ''p''<sub>2</sub>) be an element of ''W'', that is, a point in the plane such that ''p''<sub>1</sub> = ''p''<sub>2</sub>, and let ''c'' be a scalar in '''R'''. Then {{nowrap|1=''c'''''p''' = (''cp''<sub>1</sub>, ''cp''<sub>2</sub>)}}; since ''p''<sub>1</sub> = ''p''<sub>2</sub>, then ''cp''<sub>1</sub> = ''cp''<sub>2</sub>, so ''c'''''p''' is an element of ''W''.

In general, any subset of the real coordinate space '''R'''<sup>''n''</sup> that is defined by a [[homogeneous system of linear equations]] will yield a subspace.
(The equation in example I was ''z''&nbsp;=&nbsp;0, and the equation in example II was ''x''&nbsp;=&nbsp;''y''.)

=== Example III ===
Again take the field to be '''R''', but now let the vector space ''V'' be the set '''R'''<sup>'''R'''</sup> of all [[function (mathematics)|function]]s from '''R''' to '''R'''.
Let C('''R''') be the subset consisting of [[continuous function]]s.
Then C('''R''') is a subspace of '''R'''<sup>'''R'''</sup>.

''Proof:''
#We know from calculus that {{nowrap|0 ∈ C('''R''') ⊂ '''R'''<sup>'''R'''</sup>}}.
#We know from calculus that the sum of continuous functions is continuous.
#Again, we know from calculus that the product of a continuous function and a number is continuous.

=== Example IV ===
Keep the same field and vector space as before, but now consider the set Diff('''R''') of all [[differentiable function]]s.
The same sort of argument as before shows that this is a subspace too.

Examples that extend these themes are common in [[functional analysis]].