Euclidean subspace

In linear algebra, an 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. [Linear algebra, as discussed in this article, is a very well-established mathematical discipline for which there are many sources. Almost all of the material in this article can be found in Lay 2005, Meyer 2001, and Strang 2005.]

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.

Note on vectors and R^"n"

In mathematics, R^"n" denotes the set of all vectors with "n" real components::$extbf\{R\}^n\; =\; left\{(x\_1,\; x\_2,\; ldots,\; x\_n)\; :\; x\_1,x\_2,ldots,x\_n\; in\; extbf\{R\}\; ight\}$ [This equation uses set-builder notation. The same notation will be used throughout this article.] Here the word vector refers to any ordered list of numbers. Vectors can be written as either ordered tuples or as columns of numbers:: [To add to the confusion, there is also an object called a row vector, usually written ["x"₁ "x"₂ ··· "x_n"] . Some books identify ordered tuples with row vectors instead of column vectors.] 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 R³.

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 nowrap| 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. [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 0v by property 3, and therefore must contain the zero vector.)] [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, thennowrap| "b"u + "c"v is an element of "S".]

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, ifare elements of a subspace "S", andare scalars, then

:"c"₁ v₁ + "c"₂ v₂ + · · · + "c_k" v_"k"

is again an element of "S".

Geometric description

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 R³:
# The singleton set nowrap| { (0, 0, 0) } is a zero-dimensional subspace of R³.
# Any line through the origin is a one-dimensional subspace of R³.
# Any plane through the origin is a two-dimensional subspace of R³.
# The entire set R³ 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).

ystems of linear equations

The solution set to any homogeneous system of linear equations with "n" variables is a subspace of R^"n":

:

For example, the set of all vectors nowrap| ("x", "y", "z") satisfying the equations

:$x\; +\; 3y\; +\; 2z\; =\; 0\; ;;;;\; ext\{and\};;;;\; 2x\; -\; 4y\; +\; 5z\; =\; 0$

is a one-dimensional subspace of R³.

Null space of a matrix

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 R³ described above is the null space of the matrix

:

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:

:

For example, the set of all vectors nowrap| ("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 R³.

pan of vectors

In linear algebra, the system of parametric equations can be written as a single vector equation:

:

The expression on the right is called a linear combination of the vectorsnowrap| (2, 5, −1) and nowrap| (3, −4, 2). These two vectors are said to span the resulting subspace.

In general, a linear combination of vectorsis 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 v₁,...,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 v₁,...,v_"k".

; Example: The "xz"-plane in R³ 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 nowrap| (1, 0, 0) and nowrap| (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 nowrap| (1, 0, 0) and then moving some distance in the direction of nowrap| (0, 0, 1).

Column space and row space

A system of linear parametric equations can also be written as a single matrix equation:

:

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

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 R³ spanned by the three vectors nowrap| (1, 0, 0), nowrap| (0, 0, 1), and nowrap| (2, 0, 3) is just the "xz"-plane, with each point on the plane described by infinitely many different values of nowrap| "t"₁, "t"₂, "t"₃.

In general, vectors v₁,...,v_"k" are called linearly independent if

:$t\_1\; extbf\{v\}\_1\; +\; cdots\; +\; t\_k\; extbf\{v\}\_k\; ;\; e;\; u\_1\; extbf\{v\}\_1\; +\; cdots\; +\; u\_k\; extbf\{v\}\_k$

fornowrap| ("t"₁, "t"₂, ..., "t_k") ≠ ("u"₁, "u"₂, ..., "u_k"). [This definition is often stated differently: vectors v₁,...,v_"k" are linearly independent if nowrap| "t"₁v₁ + ··· + "t_k"v_"k" ≠ 0 for nowrap| ("t"₁, "t"₂, ..., "t_k") ≠ (0, 0, ..., 0). The two definitions are equivalent.] If nowrap| v₁, ..., v_"k" are linearly independent, then the coordinates nowrap| "t"₁, ..., "t_k" 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 R⁴ defined by the equations::$x\_1\; =\; 2\; x\_2;;;;\; ext\{and\};;;;x\_3\; =\; 5x\_4$:Then the vectors nowrap| (2, 1, 0, 0) and nowrap| (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 R⁴ passing through the points nowrap| (0, 0, 0, 0), nowrap| (2, 1, 0, 0), and nowrap| (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 nowrap| "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".See the article on row space for an example.

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.

ubspace membership

:Input A basis {b₁, b₂, ..., 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 b₁,...,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 nowrap| {b₁, ..., b_"k", v} are linearly dependent, and therefore nowrap| 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.See the article on column space for an example.

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 {b₁, b₂, ..., b_"k"} for a subspace "S" of R^"n", and a vector nowrap| v ∈ "S":Output Numbers "t"₁, "t"₂, ..., "t"_"k" such that nowrap|1= v = "t"₁b₁ + ··· + "t"_"k"b_"k":# Create an augmented matrix "A" whose columns are b₁,...,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 nowrap| "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 nowrap| "x"₁, "x"₂, ..., "x_n" are free. Write equations for the dependent variables in terms of the free variables.:# For each free variable "x_i", choose a vector in the null space for which nowrap|1= "x_i" = 1 and the remaining free variables are zero. The resulting collection of vectors is a basis for the null space of "A".See the article on null space for an example.

Equations for a subspace

:Input A basis {b₁, b₂, ..., 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 nowrap| b₁, b₂, ..., b_"k".:# Use elementary row operations to put "A" into reduced row echelon form.:# Let nowrap| c₁, c₂, ..., 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 c₁,...,c_"n". The nowrap| ("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

::

:then the column vectors nowrap| c₁, ..., c₆ satisfy the equations

::

:It follows that the row vectors of "A" satisfy the equations

::

:In particular, the row vectors of "A" are a basis for the null space of the corresponding matrix.

Operations on subspaces

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 [That is, the intersection of generic subspaces nowrap| "U", "V" ⊂ R^"n" has dimension nowrap| dim("U") + dim("V") − "n", or dimension zero if this number is negative.] , and the maximum only occurs when one subspace is contained in the other. For example, the intersection of two-dimensional subspaces in R³ has dimension one or two (with two only possible if they are the same plane). The intersection of three-dimensional subspaces in R⁵ has dimension one, two, or three, with most pairs intersecting along a line.

The codimension of a subspace "U" in R^"n" is the differencenowrap| "n" − dim("U"). Using codimension, the inequality above can be written

:$max(\; ext\{codim\; \}\; U,,\; ext\{codim\; \}\; V)\; leq\; ext\{codim\}(U\; cap\; V)\; leq\; ext\{codim\}(U)\; +\; ext\{codim\}(V)\; ext\{.\}$

um

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. [That is, the sum of two generic subspaces nowrap| "U", "V" − R^"n" has dimensionnowrap| dim("U") + dim("V"), or dimension "n" if this number exceeds "n".] The dimension of the intersection and the sum are related:

:$dim(U+V)\; =\; dim(U)\; +\; dim(V)\; -\; dim(U\; cap\; V)$

Orthogonal complement

The orthogonal complement of a subspace "U" is the subspace

:

Here x · u denotes the dot product of x and u. For example, if "U" is a plane through the origin in R³, then "U"^⊥ is the line perpendicular to this plane at the origin.

If b₁, b₂, ..., 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

:

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:

:

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.

ee also

* Linear algebra
* Vector space
* Linear subspace
* Flat (geometry)

Notes

References

External links

* [http://video.google.com/videoplay?docid=-584643457858917136 MIT Linear Algebra Lecture on the Four Fundamental Subspaces] at Google Video, from MIT OpenCourseWare