- Linear subspace
The concept of a linear subspace (or vector subspace) is important in
linear algebra and related fields ofmathematics .A linear subspace is usually called simply a "subspace" when the context serves to distinguish it from other kinds ofsubspace s.Definition and useful characterization
Let "K" be a field (such as the field of
real number s), and let "V" be avector space over "K".As usual, we call elements of "V" "vectors" and call elements of "K" "scalars".Suppose that "W" is asubset 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, 0, 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.Conversely, if "W" is subspace of "V", then "W" is itself a vector space under the operations induced by "V", so properties 2 and 3 are satisfied. By property 3, "-w" is in "W" whenever "w" is, and it follows that "W" is closed under subtraction as well. Since "W" is nonempty, there is an element "x" in "W", and is in "W", so property 1 is satisfied. One can also argue that since "W" is nonempty, there is an element "x" in "W", and 0 is in the field "K" so and therefore property 1 is satisfied.
Examples
Examples related to analytic geometry
Example I:Let the field "K" be the set R of
real number s, and let the vector space "V" be theEuclidean 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 R2.Take "W" to be the set of points ("x","y") of R2 such that "x" = "y".Then "W" is a subspace of R2."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 equation s 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 RR of all functions from R to R.Let C(R) be the subset consisting of continuous functions.Then C(R) is a subspace of RR.
"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 combination s.That is, "W" is a subspaceif 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".
#Since "U" and "V" are vector spaces, then 0 belongs to both sets. Thus, 0 belongs to "U" ∩ "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 theorthogonal complement of any subspace of "V" is again a subspace.External links
* [http://video.google.com/videoplay?docid=-584643457858917136 MIT Linear Algebra Lecture on the Four Fundamental Subspaces] at Google Video, from MIT OpenCourseWare
*.
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