The concept of a 'linear subspace' (or 'vector subspace') is important in
linear algebra and related fields of
mathematics.
A linear subspace is usually called simply a ''subspace'' when the context serves to distinguish it from other kinds of
subspaces.
Definition and useful characterization
Let ''K'' be a
field (such as the field of
real numbers), and let ''V'' be a
vector space over ''K''.
As usual, we call elements of ''V'' ''vectors'' and call elements of ''K'' ''scalars''.
Suppose that ''W'' is a
subset of ''V''.
If ''W'' is a vector space itself, with the same vector space operations as ''V'' has, then it is a 'subspace' of ''V''.
To use this definition, we don't have to prove that all the properties of a vector space hold for ''W''.
Instead, we can prove a theorem that gives us an easier way to show that a subset of a vector space is a subspace.
'Theorem:'
Let ''V'' be a vector space over the field ''K'', and let ''W'' be a subset of ''V''.
Then ''W'' is a subspace
if and only if it satisfies the following 3 conditions:
#The zero vector, ''θ'', is in ''W''.
#If 'u' and 'v' are elements of ''W'', then the sum 'u' + 'v' is an element of ''W'';
#If 'u' is an element of ''W'' and ''c'' is a scalar from ''K'', then the scalar product ''c'''u' is an element of ''W'';
'Proof:'
Firstly, property 1 ensures ''W'' is nonempty. Looking at the definition of a
vector space, we see that properties 2 and 3 above assure closure of ''W'' under addition and scalar multiplication, so the vector space operations are well defined. Since elements of ''W'' are necessarily elements of ''V'', axioms 1, 2 and 5-8 of a vector space are satisfied ''
a fortiori''. By the closure of ''W'' under scalar multiplication (specifically by 0 and -1), axioms 3 and 4 of a vector space are satisfied.
Vector subspaces are
subgroups of vector spaces under
vector addition.
Examples
Examples related to analytic geometry
'Example I:'
Let the field ''K'' be the set 'R' of
real numbers, and let the vector space ''V'' be the
Euclidean space 'R'
''3''.
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 'u' = (''u''
1,''u''
2,0) and 'v' = (''v''
1,''v''
2,0). Then 'u' + 'v' = (''u''
1+''v''
1,''u''
2+''v''
2,0+0) = (''u''
1+''v''
1,''u''
2+''v''
2,0). Thus, 'u' + 'v' is an element of ''W'', too.
#Given 'u' in ''W'' and a scalar ''c'' in 'R', if 'u' = (''u''
1,''u''
2,0) again, then ''c'''u' = (''cu''
1, ''cu''
2, ''c''0) = (''cu''
1,''cu''
2,0). Thus, ''c'''u' is an element of ''W'' too.
'Example II:'
Let the field be 'R' again, but now let the vector space be the
Euclidean geometry 'R'
2.
Take ''W'' to be the set of points (''x'',''y'') of 'R'
2 such that ''x'' = ''y''.
Then ''W'' is a subspace of 'R'
2.
''Proof:''
#Let 'p' = (''p''
1,''p''
2) and 'q' = (''q''
1,''q''
2) be elements of ''W'', that is, points in the plane such that ''p''
1 = ''p''
2 and ''q''
1 = ''q''
2. Then 'p' + 'q' = (''p''
1+''q''
1,''p''
2+''q''
2); since ''p''
1 = ''p''
2 and ''q''
1 = ''q''
2, then ''p''
1 + ''q''
1 = ''p''
2 + ''q''
2, so 'p' + 'q' is an element of ''W''.
#Let 'p' = (''p''
1,''p''
2) be an element of ''W'', that is, a point in the plane such that ''p''
1 = ''p''
2, and let ''c'' be a scalar in 'R'. Then ''c'''p' = (''cp''
1,''cp''
2); since ''p''
1 = ''p''
2, then ''cp''
1 = ''cp''
2, so ''c'''p' is an element of ''W''.
In general, any subset of an Euclidean space 'R'
''n'' that is defined by a system of homogeneous
linear equations will yield a subspace.
(The equation in example I was ''z'' = 0, and the equation in example II was ''x'' = ''y''.)
Geometrically, these subspaces are points, lines, planes, and so on, that pass through the point '0'.
Examples related to calculus
'Example III:'
Again take the field to be 'R', but now let the vector space ''V'' be the set 'R'
'R' of all
functions from 'R' to 'R'.
Let C('R') be the subset consisting of
continuous functions.
Then C('R') is a subspace of 'R'
'R'.
''Proof:''
#We know from calculus 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 functions.
The same sort of argument as before shows that this is a subspace too.
Examples that extend these themes are common in
functional analysis.
Properties of subspaces
A way to characterise subspaces is that they are closed under
linear combinations.
That is, ''W'' is a subspace
if and only if every linear combination of (
finitely many) elements of ''W'' also belongs to ''W''.
Conditions 1 and 2 for a subspace are simply the most basic kinds of linear combinations.
Operations on subspaces
Given subspaces ''U'' and ''W'' of a vector space ''V'', then their
intersection ''U'' ∩ ''W'' := {'v' ∈ ''V'' : 'v' is an element of both ''U'' and ''W''} is also a subspace of ''V''.
''Proof:''
#Let 'v' and 'w' be elements of ''U'' ∩ ''W''. Then 'v' and 'w' belong to both ''U'' and ''W''. Because ''U'' is a subspace, then 'v' + 'w' belongs to ''U''. Similarly, since ''W'' is a subspace, then 'v' + 'w' belongs to ''W''. Thus, 'v' + 'w' belongs to ''U'' ∩ ''W''.
#Let 'v' belong to ''U'' ∩ ''W'', and let ''c'' be a scalar. Then 'v' belongs to both ''U'' and ''W''. Since ''U'' and ''W'' are subspaces, ''c'''v' belongs to both ''U'' and ''W''.
Furthermore, the sum
:
is also a subspace of ''V''. The
dimensions of ''U'' ∩ ''W'' and ''U'' + ''W'' satisfy
:
For every vector space ''V'', the set {'0'} and ''V'' itself are subspaces of ''V''.
If ''V'' is an
inner product space, then the
orthogonal complement of any subspace of ''V'' is again a subspace.
External links
★
MIT Linear Algebra Lecture on the Four Fundamental Subspaces at Google Video, from MIT OpenCourseWare
★ .
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