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Description of vector spaces, subspaces, bases and spans.

Vector spaces

Definition 1 A linear vector space ( X , R , + , · ) is given by a signal space X (called vectors), a set of scalars R , an addition operation + : X × X X , and a multiplication operation · : R × X X , such that:

  1. X forms a group under addition:
    1. x , y X ! x + y X , (closed under addition)
    2. 0 X such that 0 + X = X + 0 = X . (additive identity)
    3. x X y X such that x + y = 0 , (additive inverse)
    4. x , y , z X x + ( y + z ) = ( x + y ) + z . (associative law)
  2. Multiplication has the following properties: for any x , y X and a , b R :
    1. a · x X , (closure in X under multiplication)
    2. a · ( b · x ) = ( a · b ) · x , (compatibility)
    3. ( a + b ) · x = a · x + b · x , (distributive law over R )
    4. a · ( x + y ) = a · x + a · y . (distributive law over X )
  3. The set R has the following properties:
    1. There exists 1 R s.t. 1 · x = x x X , (multiplication identity)
    2. There exists 0 R s.t. 0 · x = 0 x X . (multiplicative null element)

Example 1 Here are some examples of vector spaces:

  • X = R n (space of all vectors of length n ) over R = R is a vector space.
  • X = C n ( C is complex numbers) over R = C is a vector space.
  • X = R n over R = C is not a vector space, because closure in X under multiplication is not met.
  • X = C [ T ] (continuous functions in T ) over R = R is a vector space.

Subspaces

Definition 2 A subset M X is a linear subspace of X if M itself is a linear vector space. Note that, in particular, this implies that any subspace M must obey 0 M .

Example 2 Here are some examples of subspaces:

  • In X = R 2 over R = R , any line that passes through the origin is a subspace of X :
    M = ( x , y ) R 2 such that y x = c .
  • In X = C [ T ] over R = R , the followings are subspaces of X :
    M 1 = { f ( x ) = a x 2 + b x + c : a , b , c R } M 2 = { f ( x ) : f ( x 0 ) = 0 } .
    In contrast, the set M 3 = { f ( x ) : f ( x 0 ) = a 0 } is not a subspace.

Proposition 1 If M and N are subspaces of X , then M N is also a subspace.

Proof: We assume that M and N hold properties of linear vector space, and show that so does M N :

  1. x , y M N x , y M x + y M x , y N x + y N x + y M N
  2. M linear vector space 0 M N linear vector space 0 N 0 M N
  3. x M N x M y M s.t. x + y = 0 x N y N s.t. x + y = 0 y M N

The other properties are shown in a similar fashion.

Definition 3 A vector x X , where ( X , R , + , · ) is a vector space, is a linear combination of a set { x 1 , x 2 , ... , x n } X if it can be written as x = i = 1 n a i · x i , a i R . The set of all linear combinations of a set of points { x 1 , x 2 , ... , x n } builds a linear subspace of X .

Example 3 Q = i = 0 2 a i x i is a linear subspace of ( C [ T ] , R , + , · ) containing the set of all quadratic functions, as it corresponds to all linear combinations of the set of functions { x 2 , x , 1 } .

Bases and spans

Definition 4 For the set S = { x 1 , x 2 , ... , x n } X , the span of S is written as

[ S ] = span ( S ) = x : x = i = 1 n a i x i , a i R .

Example 4 The space of quadratic functions Q is written as Q = [ S 1 ] , with S 1 = { x 2 , x , 1 } . The space can also be written as [ S 2 ] with S 2 = { 1 , x , x 2 - 2 } ) , i.e. [ S 1 ] = [ S 2 ] . To prove this, we need to show [ S 2 ] [ S 1 ] and [ S 1 ] [ S 2 ] . For the former case we have

x = a 1 + a 2 x + a 3 ( x 2 - 2 ) = ( a 1 - 2 a 3 ) + a 2 x + a 3 x 2 ,

which means that every element that can be spanned by S 2 , can also be spanned by S 1 , and hence [ S 2 ] [ S 1 ] . The latter case can be shown in a similar manner.

Definition 5 A set S is a linearly independent set if

i = 1 n a i x i = 0 a i = 0 , i { 1 , 2 , ... , n } .

Otherwise, the set S is linearly dependent .

Definition 6 A finite set S of linearly independent vectors is a basis for the space X if [ S ] = X , i.e. if X is spanned by S .

Definition 7 The dimension of X is the number of elements of its basis | S | . A vector space for which a finite basis does not exist is called an infinite-dimensional space .

Theorem 1 Any two bases of a subspace have the same number of elements.

Proof: We prove by contradiction: assume that S 1 = { x 1 , ... , x n } and S 2 = { y 1 , ... , y m } , m > n , are two bases for a subspace X with different numbers of elements. We have that since y 1 X it can be written as a linear combination of S 1 :

y 1 = i = 1 n a i x i .

Order the elements of S 1 above so that a 1 is nonzero; since y 1 must be nonzero then at least one such a i must exist. Solving the above equation for x 1 yields

x 1 = 1 a 1 y 1 - i = 2 n a i x i .

Thus { y 1 , x 2 , x 3 , ... , x n } is a basis, in terms of which we can write any vector of the space X , including y 2 :

y 2 = b 1 y 1 + i = 2 n b i x i .

Since y 1 , y 2 are linearly independent, at least one of the values of b i , i > 2 , must be nonzero. Sort the remaining x i so that b 2 is nonzero. Solving for x 2 results in

x 2 = 1 b 2 y 2 - b 1 b 2 y 1 + i = 3 n b i b 2 x i .

Therefore, { y 1 , y 2 , x 3 , ... , x n } is a basis for X . Continuing in this way, we can eliminate each x i , showing that { y 1 , y 2 , ... , y n } is a basis for X . Thus, we have y n + 1 = i = 1 n c i y i , or equivalently:

c n + 1 y n + 1 + i = 1 n c i y i = 0 with c n + 1 = - 1 .

As a result, S 2 is linearly dependent and is not a basis. Therefore, all bases of X must have the same number of elements.

Basis representations

Having a basis in hand for a given subspace allows us to express the points in the subspace in more than one way. For each point x [ S ] in the span of a basis S = { S 1 , S 2 ... S n } , that is,

x = i = 1 n a i S i ,

there is a one-to-one map (i.e., an equivalence) between x [ S ] and a = { a 1 , ... , a n } R n , that is, both x and a uniquely identify the point in S . This is stated more formally as a theorem.

Theorem 2 If S is a linearly independent set, then

i = 1 n a i S i = i = 1 n b i S i

if and only if a i = b i for i = 1 , 2 ... n .

Proof: Theorem 1 states that the scalars { a 1 , ... , a n } are unique for x . We begin by assuming that indeed

i = 1 n a i S i = i = 1 n b i S i .

This implies

i = 1 n a i S i - i = 1 n b i S i = 0 , i = 1 n ( a i - b i ) S i = 0 .

Since the elements of S are linearly independent, each one of the scalars of the sum must be zero, that is, a i - b i = 0 and so a i = b i for each i = 1 , ... , n .

Example 5 (Digital Communications) A transmitter sends two waveforms:

S 1 ( t ) = 2 / T cos ( 2 π f c t ) t [ 0 , T ] if bit 1 is transmitted,
S 0 ( t ) = 2 / T sin ( 2 π f c t ) t [ 0 , T ] if bit 0 is transmitted.

The signal r ( t ) recorded by the receiver is continuous, that is, r ( t ) C [ T ] . Assuming that the propagation delay is known and corrected at the receiver, we will have they the received signal must be in the span of the possible transmitted signals, i.e., r ( t ) span ( S 1 ( t ) , S 0 ( t ) ) . One can check that S 1 ( t ) and S 2 ( t ) are linearly independent. Thus, one can use a unique choice of coefficients a 0 and a 1 that denote whether bit 0 or bit 1 is transmitted and contain the amount of attenuation caused by the transmission:

r ( t ) = a 1 S 1 ( t ) + a 0 S 0 ( t ) .

The uniqueness of this representation can only be obtained if the transmitted signals S 0 ( t ) and S 1 ( t ) are linearly independent. The waveforms above are used in in phase shift keying (PSK); other similar examples include frequency shift keying (FSK) and quadrature amplitude modulation (QAM).

Questions & Answers

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Source:  OpenStax, Introduction to compressive sensing. OpenStax CNX. Mar 12, 2015 Download for free at http://legacy.cnx.org/content/col11355/1.4
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