EUCLIDEAN SUBSPACE

Three one-dimensional subspaces of 'R'2

In linear algebra, a 'Euclidean subspace' (or 'subspace of R''n''') is a set of vectors that is closed under addition and scalar multiplication. Geometrically, a subspace is a flat in ''n''-dimensional Euclidean space that passes through the origin. Examples of subspaces include the solution set to a homogeneous system of linear equations, the subset of Euclidean space described by a system of homogeneous linear parametric equations, the span of a collection of vectors, and the null space, column space, and row space of a matrix.
In abstract linear algebra, Euclidean subspaces are important examples of vector spaces. In this context, a Euclidean subspace is simply a linear subspace of a Euclidean space.

Contents
Note on vectors and 'R'''n''
Definition
Geometric description
Systems of linear equations
Null space of a matrix
Linear parametric equations
Span of vectors
Column space and row space
Independence, basis, and dimension
Algorithms
Basis for a row space
Subspace membership
Basis for a column space
Coordinates for a vector
Basis for a null space
Equations for a subspace
Operations on subspaces
Intersection
Sum
Orthogonal complement
See also
Notes
References
External links

Note on vectors and 'R'''n''


In mathematics, 'R'''n'' denotes the set of all vectors with ''n'' real components:
: extbf{R}^n = left{(x_1, x_2, ldots, x_n) : x_1,x_2,ldots,x_n in extbf{R}
ight}[1]
Here the word 'vector' refers to any ordered list of numbers. Vectors can be written as either ordered tuples or as columns of numbers:
:(x_1, x_2, ldots, x_n) = left[!! egin{array}{c} x_1 \ x_2 \ dots \ x_n end{array} !!
ight][2]
Geometrically, we regard vectors with ''n'' components as points in an ''n''-dimensional space. That is, we identify the set 'R'''n'' with ''n''-dimensional Euclidean space. Any subset of 'R'''n'' can be thought of as a geometric object (namely the object consisting of all the points in the subset). Using this mode of thought, a line in three-dimensional space is the same as the set of points on the line, and is therefore just a subset of 'R'3.

Definition


A 'Euclidean subspace' is a subset ''S'' of 'R'''n'' with the following properties:
# The zero vector '0' is an element of ''S''.
# If 'u' and 'v' are elements of ''S'', then 'u' + 'v' is an element of ''S''.
# If 'v' is an element of ''S'' and ''c'' is a scalar, then ''c'''v' is an element of ''S''.
There are several common variations on these requirements, all of which are logically equivalent to the list above.[3]
[4]
Because subspaces are closed under both addition and scalar multiplication, any linear combination of vectors from a subspace is again in the subspace. That is, if 'v'1, 'v'2, ..., 'v'''k'' are elements of a subspace ''S'', and ''c''1, ''c''2, ..., ''ck'' are scalars, then
:''c''1 'v'1 + ''c''2 'v'2 + · · · ''ck'' 'v'''k''
is again an element of ''S''.

Geometric description


Three two-dimensional subspaces of 'R'3. The center point is the zero vector.

Geometrically, a subspace of 'R'''n'' is simply a flat through the origin, i.e. a copy of a lower dimensional (or equi-dimensional) Euclidean space sitting in ''n'' dimensions. For example, there are four different types of subspaces in 'R'3:
# The singleton set { (0, 0, 0) } is a 'zero-dimensional' subspace of 'R'3.
# Any line through the origin is a 'one-dimensional' subspace of 'R'3.
# Any plane through the origin is a 'two-dimensional' subspace of 'R'3.
# The entire set 'R'3 is a 'three-dimensional' subspace of itself.
In ''n''-dimensional space, there are subspaces of every dimension from 0 to ''n''.
The geometric dimension of a subspace is the same as the number of vectors required for a basis (see below).

Systems of linear equations


The solution set to any homogeneous system of linear equations with ''n'' variables is a subspace of 'R'''n'':
:left{ left[!! egin{array}{c} x_1 \ x_2 \ dots \ x_n end{array} !!
ight] in extbf{R}^n : egin{alignat}{6}
a_{11} x_1 &&; + ;&& a_{12} x_2 &&; + cdots + ;&& a_{1n} x_n &&; = 0& \
a_{21} x_1 &&; + ;&& a_{22} x_2 &&; + cdots + ;&& a_{2n} x_n &&; = 0& \
dots;;; && && dots;;; && && dots;;; && dots,& \
a_{m1} x_1 &&; + ;&& a_{m2} x_2 &&; + cdots + ;&& a_{mn} x_n &&; = 0&
end{alignat}
ight}
For example, the set of all vectors (''x'', ''y'', ''z'') satisfying the equations
:x+3y+2z=0;;;; ext{and};;;;2x-4y+5z=0
is a one-dimensional subspace of 'R'3.
Null space of a matrix

Main articles: Null space

In linear algebra, a homogeneous system of linear equations can be written as a single matrix equation:
:A extbf{x} = extbf{0}
The set of solutions to this equation is known as the null space of the matrix. For example, the subspace of 'R'3 described above is the null space of the matrix
:A = left[ egin{alignat}{3} 1 && 3 && 2 &\ 2 && ;;-4 && ;;;;5 &end{alignat} ,
ight] ext{.}
Every subspace of 'R'''n'' can be described as the null space of some matrix (see algorithms, below).

Linear parametric equations


The subset of 'R'''n'' described by a system of homogeneous linear parametric equations is a subspace:
:left{ left[!! egin{array}{c} x_1 \ x_2 \ dots \ x_n end{array} !!
ight] in extbf{R}^n : egin{alignat}{7}
x_1 &&; = ;&& a_{11} t_1 &&; + ;&& a_{12} t_2 &&; + cdots + ;&& a_{1m} t_m & \
x_2 &&; = ;&& a_{21} t_1 &&; + ;&& a_{22} t_2 &&; + cdots + ;&& a_{2m} t_m & \
dots ,&& && dots;;; && && dots;;; && && dots;;; & \
x_n &&; = ;&& a_{n1} t_1 &&; + ;&& a_{n2} t_2 &&; + cdots + ;&& a_{nm} t_m & \
end{alignat} ext{ for some } t_1,ldots,t_min extbf{R}
ight}
For example, the set of all vectors (''x'', ''y'', ''z'') parameterized by the equations
:x=2t_1+3t_2,;;;;y=5t_1-4t_2,;;;; ext{and};;;;z=-t_1+2t_2
is a two-dimensional subspace of 'R'3.
Span of vectors

Main articles: Linear span

In linear algebra, the system of parametric equations can be written as a single vector equation:
:left[ egin{alignat}{1} x& \ y& \ z& end{alignat},
ight] ;=; t_1 !left[ egin{alignat}{1} 2& \ 5& \ -1& end{alignat},
ight] + t_2 !left[ egin{alignat}{1} 3& \ -4& \ 2& end{alignat},
ight]
The expression on the right is called a linear combination of the vectors (2, 5, –1) and (3, –4, 2). These two vectors are said to 'span' the resulting subspace.
In general, a 'linear combination' of vectors 'v'1, 'v'2, . . . , 'v'''k'' is any vector of the form
:t_1 extbf{v}_1 + cdots + t_k extbf{v}_k ext{.}
The set of all possible linear combinations is called the 'span':
: ext{Span} { extbf{v}_1, ldots, extbf{v}_k } = left{ t_1 extbf{v}_1 + cdots + t_k extbf{v}_k : t_1,ldots,t_kinmathbf{R}
ight}
If the vectors 'v'1,...,'v'''k'' have ''n'' components, then their span is a subspace of 'R'''n''. Geometrically, the span is the flat through the origin in ''n''-dimensional space determined by the points 'v'1,...,'v'''k''.
; Example
: The ''xz''-plane in 'R'3 can be parameterized by the equations
::x=t_1,;;;y=0,;;;z=t_2
:As a subspace, the ''xz''-plane is spanned by the vectors (1, 0, 0) and (0, 0, 1). Every vector in the ''xz''-plane can be written as a linear combination of these two:
::(t_1, 0, t_2) = t_1(1,0,0) + t_2(0,0,1) ext{.},
:Geometrically, this corresponds to the fact that every point on the ''xz''-plane can be reached from the origin by first moving some distance in the direction of (1, 0, 0) and then moving some distance in the direction of (0, 0, 1).
Column space and row space

Main articles: Row and column spaces

A system of linear parametric equations can also be written as a single matrix equation:
: extbf{x} = A extbf{t};;;; ext{where};;;;A = left[ egin{alignat}{2} 2 && 3 & \ 5 && ;;-4 & \ -1 && 2 & end{alignat} ,
ight] ext{.}
In this case, the subspace consists of all possible values of the vector 'x'. In linear algebra, this subspace is known as the 'column space' (or image) of the matrix ''A''. It is precisely the subspace of 'R'''n'' spanned by the column vectors of ''A''.
The 'row space' of a matrix is the subspace spanned by its row vectors. The row space is interesting because it is the orthogonal complement of the null space (see below).
Independence, basis, and dimension

Main articles: Linear independence, Basis (linear algebra), Dimension (vector space)

The vectors 'u' and 'v' are a basis for this two-dimensional subspace of 'R'3.

In general, a subspace of 'R'''n'' determined by ''k'' parameters (or spanned by ''k'' vectors) has dimension ''k''. However, there are exceptions to this rule. For example, the subspace of 'R'3 spanned by the three vectors (1, 0, 0), (0, 0, 1), and (2, 0, 3) is just the ''xz''-plane, with each point on the plane described by infinitely many different values of ''t''1, ''t''2, ''t''3.
In general, vectors 'v'1,...,'v'''k'' are called 'linearly independent' if
:t_1 extbf{v}_1 + cdots + t_k extbf{v}_k ;
e; u_1 extbf{v}_1 + cdots + u_k extbf{v}_k
for (''t''1, ''t''2, ..., ''tk'') ≠ (''u''1, ''u''2, ..., ''uk'').[5] If 'v'1, ..., 'v'''k'' are linearly independent, then the 'coordinates' ''t''1, ..., ''tk'' for a vector in the span are uniquely determined.
A 'basis' for a subspace ''S'' is a set of linearly independent vectors whose span is ''S''. The number of elements in a basis is always equal to the geometric dimension of the subspace. Any spanning set for a subspace can be changed into a basis by removing redundant vectors (see algorithms, below).
; Example
: Let ''S'' be the subspace of 'R'4 defined by the equations
::x_1 = 2 x_2;;;; ext{and};;;;x_3 = 5x_4
:Then the vectors (2, 1, 0, 0) and (0, 0, 5, 1) are a basis for ''S''. In particular, every vector that satisfies the above equations can be written uniquely as a linear combination of the two basis vectors:
::(2t_1, t_1, 5t_2, t_2) = t_1(2, 1, 0, 0) + t_2(0, 0, 5, 1),
:The subspace ''S'' is two-dimensional. Geometrically, it is the plane in 'R'4 passing through the points (0, 0, 0, 0), (2, 1, 0, 0), and (0, 0, 5, 1).

Algorithms


Most algorithms for dealing with subspaces involve row reduction. This is the process of applying elementary row operations to a matrix until it reaches either row echelon form or reduced row echelon form. Row reduction has the following important properties:
# The reduced matrix has the same null space as the original.
# Row reduction does not change the span of the row vectors, i.e. the reduced matrix has the same row space as the original.
# Row reduction does not affect the linear dependence of the column vectors.
Basis for a row space

:'Input' An ''m'' × ''n'' matrix ''A''.
:'Output' A basis for the row space of ''A''.
:# Use elementary row operations to put ''A'' into row echelon form.
:# The nonzero rows of the echelon form are a basis for the row space of ''A''.
If we instead put the matrix ''A'' into reduced row echelon form, then the resulting basis for the row space is uniquely determined. This provides an algorithm for checking whether two row spaces are equal and, by extension, whether two subspaces of 'R'''n'' are equal.
Subspace membership

:'Input' A basis {'b'1, 'b'2, ..., 'b'''k''} for a subspace ''S'' of 'R'''n'', and a vector 'v' with ''n'' components.
:'Output' Determines whether 'v' is an element of ''S''
:# Create a (''k'' + 1) × ''n'' matrix ''A'' whose rows are the vectors 'b'1,...,'b'''k'' and 'v'.
:# Use elementary row operations to put ''A'' into row echelon form.
:# If the echelon form has a row of zeroes, then the vectors {'b'1, ..., 'b'''k'', 'v'} are linearly dependent, and therefore 'v' ∈ ''S''.
Basis for a column space

:'Input' An ''m'' × ''n'' matrix ''A''
:'Output' A basis for the column space of ''A''
:# Use elementary row operations to put ''A'' into row echelon form.
:# Determine which columns of the echelon form have pivots. The corresponding columns of the original matrix are a basis for the column space.
This produces a basis for the column space that is a subset of the original column vectors. It works because the columns with pivots are a basis for the column space of the echelon form, and row reduction does not change the linear dependence relationships between the columns.
Coordinates for a vector

:'Input' A basis {'b'1, 'b'2, ..., 'b'''k''} for a subspace ''S'' of 'R'''n'', and a vector 'v' ∈ ''S''
:'Output' Numbers ''t''1, ''t''2, ..., ''t''''k'' such that 'v' = ''t''1'b'1 + ··· + ''t''''k'''b'''k''
:# Create an augmented matrix ''A'' whose columns are 'b'1,...,'b'''k'' , with the last column being 'v'.
:# Use elementary row operations to put ''A'' into reduced row echelon form.
:# Express the final column of the reduced echelon form as a linear combination of the first ''k'' columns. The coefficients used are the desired numbers ''t''1, ''t''2, ..., ''t''''k''. (These should be precisely the first ''k'' entries in the final column of the reduced echelon form.)
If the final column of the reduced row echelon form contains a pivot, then the input vector 'v' does not lie in ''S''.
Basis for a null space

:'Input' An ''m'' × ''n'' matrix ''A''.
:'Output' A basis for the null space of ''A''
:# Use elementary row operations to put ''A'' in reduced row echelon form.
:# Using the reduced row echelon form, determine which of the variables ''x''1, ''x''2, ..., ''xn'' are free. Write equations for the dependent variables in terms of the free variables.
:# For each free variable ''xi'', choose a vector in the null space for which ''xi'' = 1 and the remaining free variables are zero. The resulting collection of vectors is a basis for the null space of ''A''.
;Example
:If the reduced row echelon form of ''A'' is
::left[ egin{alignat}{6}
1 && 0 && -3 && 0 && 2 && 0 \
0 && 1 && 5 && 0 && -1 && 4 \
0 && 0 && 0 && 1 && 7 && -9 \
0 && ;;;;;0 && ;;;;;0 && ;;;;;0 && ;;;;;0 && ;;;;;0 end{alignat} ,
ight]
:then ''x''3, ''x''5, and ''x''6 are free, with ''x''1 = 3''x''3 – 2''x''5,  ''x''2 = –5''x''3 + ''x''5 – 4''x''6, and ''x''4 = –7''x''5 + 9''x''6. Therefore,
:: left[!! egin{array}{r} 3 \ -5 \ mathbf{1} \ 0 \ mathbf{0} \ mathbf{0} end{array}
ight],;
left[!! egin{array}{r} -2 \ 1 \ mathbf{0} \ -7 \ mathbf{1} \ mathbf{0} end{array}
ight],;
left[!! egin{array}{r} 0 \ -4 \ mathbf{0} \ 9 \ mathbf{0} \ mathbf{1} end{array}
ight]
:is a basis for the null space.
Equations for a subspace

:'Input' A basis {'b'1, 'b'2, ..., 'b'''k''} for a subspace ''S'' of 'R'''n''
:'Output' An (''n'' – ''k'') × ''n'' matrix whose null space is ''S''.
:# Create a matrix ''A'' whose rows are 'b'1, 'b'2, ..., 'b'''k''.
:# Use elementary row operations to put ''A'' into reduced row echelon form.
:# Let 'c'1, 'c'2, ..., 'c'''n'' be the columns of the reduced row echelon form. For each column without a pivot, write an equation expressing the column as a linear combination of the columns with pivots.
:# This results in a homogeneous system of ''n'' – ''k'' linear equations involving the variables 'c'1,...,'c'''n''. The (''n'' – ''k'') × ''n'' matrix corresponding to this system is the desired matrix with nullspace ''S''.
; Example
:If the reduced row echelon form of ''A'' is
::left[ egin{alignat}{6}
1 && 0 && -3 && 0 && 2 && 0 \
0 && 1 && 5 && 0 && -1 && 4 \
0 && 0 && 0 && 1 && 7 && -9 \
0 && ;;;;;0 && ;;;;;0 && ;;;;;0 && ;;;;;0 && ;;;;;0 end{alignat} ,
ight]
:then the column vectors 'c'1, ..., 'c'6 satisfy the equations
:: egin{alignat}{1}
extbf{c}_3 &= -3 extbf{c}_1 + 5 extbf{c}_2 \
extbf{c}_5 &= 2 extbf{c}_1 - extbf{c}_2 + 7 extbf{c}_3 \
extbf{c}_6 &= 4 extbf{c}_2 - 9 extbf{c}_3
end{alignat} ext{.}
:It follows that the row vectors of ''A'' satisfy the equations
:: egin{alignat}{1}
x_3 &= -3x_1 + 5x_2 \
x_5 &= 2x_1 - x_2 + 7x_3 \
x_6 &= 4x_2 - 9x_3
end{alignat} ext{.}
:In particular, the row vectors of ''A'' are a basis for the null space of the corresponding matrix.

Operations on subspaces


In 'R'3, the intersection of two-dimensional subspaces is one-dimensional.

Intersection

If ''U'' and ''V'' are subspaces of 'R'''n'', their intersection is also a subspace:
:U cap V = left{ extbf{x}in extbf{R}^n : extbf{x}in U ext{ and } extbf{x}in V
ight}
The dimension of the intersection satisfies the inequality
:dim(U) + dim(V) - n leq dim(U cap V) leq min(dim U,,dim V) ext{.}
The minimum is the most general case[6], and the maximum only occurs when one subspace is contained in the other. For example, the intersection of two-dimensional subspaces in 'R'3 has dimension one or two (with two only possible if they are the same plane). The intersection of three-dimensional subspaces in 'R'5 has dimension one, two, or three, with most pairs intersecting along a line.
The 'codimension' of a subspace ''U'' in 'R'''n'' is the difference ''n'' – dim(''U''). Using codimension, the inequality above can be written
:max( ext{codim } U,, ext{codim } V) leq ext{codim}(U cap V) leq ext{codim}(U) + ext{codim}(V) ext{.}
Sum

If ''U'' and ''V'' are subspaces of 'R'''n'', their 'sum' is the subspace
:U + V = left{ extbf{u} + extbf{v} : extbf{u}in U ext{ and } extbf{v}in V
ight} ext{.}
For example, the sum of two lines is the plane that contains them both. The dimension of the sum satisfies the inequality
:max(dim U,dim V) leq dim(U + V) leq dim(U) + dim(V) ext{.}
Here the minimum only occurs if one subspace is contained in the other, while the maximum is the most general case.[7] The dimension of the intersection and the sum are related:
:dim(U+V) = dim(U) + dim(V) - dim(U cap V)
Orthogonal complement

Main articles: Orthogonal complement

The 'orthogonal complement' of a subspace ''U'' is the subspace
:U^ot = left{ extbf{x}in extbf{R}^n : extbf{x} cdot extbf{u}=0 ext{ for every } extbf{u}in U
ight}
Here 'x' · 'u' denotes the dot product of 'x' and 'u'. For example, if ''U'' is a plane through the origin in 'R'3, then ''U'' is the line perpendicular to this plane at the origin.
If 'b'1, 'b'2, ..., 'b'''k'' is a basis for ''U'', then a vector 'x' is in the orthogonal complement of ''U'' if and only if it is orthogonal to each 'b'''i''. It follows that the null space of a matrix is the orthogonal complement of the row space.
The dimension of a subspace and its orthogonal complement are related by the equation
:dim(U) + dim(U^ot) = n
That is, the dimension of ''U'' is equal to the codimension of ''U''. The intersection of ''U'' and ''U'' is the origin, and the sum of ''U'' and ''U'' is all of 'R'''n''
Orthogonal complements satisfy a version of De Morgan's laws:
:(U + V)^ot = U^ot cap V^ot;;;; ext{and};;;;(U cap V)^ot = U^ot + V^ot ext{.}
In fact, the collection of subspaces of 'R'''n'' satisfy all of the axioms for a Boolean algebra, with intersection as AND, sum as OR, and orthogonal complement as NOT.

See also



Linear algebra

Euclidean space

n-dimensional space

Vector space

Linear subspace

Flat (geometry)

System of linear equations

Null space

Row and column spaces

Linear span

Linear independence

Basis (linear algebra)

Dimension (vector space)

Row reduction

Orthogonal complement

Notes


1. This equation uses set-builder notation. The same notation will be used throughout this article.
2. To add to the confusion, there is also an object called a row vector, usually written [''x''1  ''x''2  ···  ''xn'']. Some books identify ordered tuples with row vectors instead of column vectors.
3. The requirement that ''S'' contains the zero vector is equivalent to requiring that ''S'' is nonempty. (Once ''S'' contains any single vector 'v' it must contain 0'v' by property 3, and therefore must contain the zero vector.)
4. The second and third requirements can be combined into the following statement: If 'u' and 'v' are elements of ''S'' and ''b'' and ''c'' are scalars, then ''b'''u' + ''c'''v' is an element of ''S''.
5. This definition is often stated differently: vectors 'v'1,...,'v'''k'' are linearly independent if ''t''1'v'1 + ··· + ''tk'''v'''k'' ≠ '0' for (''t''1, ''t''2, ..., ''tk'') ≠ (0, 0, ..., 0). The two definitions are equivalent.
6. That is, the intersection of generic subspaces ''U'', ''V'' ⊂ 'R'''n'' has dimension dim(''U'') + dim(''V'') – ''n'', or dimension zero if this number is negative.
7. That is, the sum of two generic subspaces ''U'', ''V'' ⊂ 'R'''n'' has dimension dim(''U'') + dim(''V''), or dimension ''n'' if this number exceeds ''n''.

References



Linear algebra and its applications, Lay, David C., , , Pearson/Addison-Wesley, 2006,

Schaum's outline of theory and problems of linear algebra, Lipson, Marc; Lipschutz, Seymour, , , McGraw-Hill, 2001,

★ Beezer, Rob, ''A First Course in Linear Algebra'', licensed under GFDL.

★ Jim Hefferon: ''Linear Algebra'' (Online textbook)

★ Edwin H. Connell: ''Elements of Abstract and Linear Algebra'' (Online textbook)

External links



MIT Linear Algebra Lecture on the Four Fundamental Subspaces at Google Video, from MIT OpenCourseWare

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