Subspaces

Author :  Thomas Shores
University of Nebraska
Send comments to :  tshores@math . unl . edu

Review

Let us recall the definition of an (abstract) vector space:

Definition:  A  VECTOR SPACE (v.s.) over a field of scalars is a set V of elements called vectors, together with two binary operations, scalar multiplication * (usually suppressed) and vector addition + , subject to the following vector space laws:   For all  u, v, w   ∈  V  and scalars  a  and b, 
(1) (Closure of vector additon)  u  +  v  ∈  V.
(2) (Commutativity of addition)  u + v  =  v + u.
(3) (Associativity of addition)  u + (v + w)  =  (u + v) + w.
(4) (Additivive Identity)  There exists an element  0 ∈ V such that  u + 0 = u = 0 + u.
(5) (Additive Inverse)  There exists, for each  u ∈ V, an element -u ∈ in V  such that
	u + (-u)  =  0  =  (-u) + u.
(6) (Closure of scalar multiplication)  a*v  ∈  V.
(7) (Distributive Law)  a*(u+v)  =  a*u  +  a*v.
(8) (Distributive Law)  (a + b)*u  =  a*u  +  b*u.
(9) (Associative Law)  (a b) * u  = a*(b*u).
(10) (Monoidal Law)   1*u  =  u.

NOTATION:  u  +  (-v)  =  u - v.
ALSO:  we often refer to the “vector space  V”, without identifying explicitly the binary operations.  They are taken for granted.
	
SIMPLE FACTS: (deducible from the vector space properties alone):
	For all vectors  u,  0*u = 0,  where the latter 0  means the zero vector.
	For all vectors  u,  (-1)*u  = -u.

EXAMPLES OF VECTOR SPACES:  
(1)  The spaces  Rn  and  Cn,   n >=1, over the fields R  and C, respectively.  Also, C is a vector space over R.
(2)  The set  M(2,2)  of all 2x2  real matrices, together with the usual matrix addition and scalar multiplication.
(3)  The set 
	V = C[0,1] = {f(x) | f(x) is a continuous function on the interval [0,1] },  together with the usual function addition  and scalar multiplication:
		(f+g)(x)  =  f(x) + g(x)
		(a*f)(x)  = a f(x).
	This is a very non-standard vector space.  It seems kind of odd, but we are really thinking of  functions  f(x),  0 <= x <= 1,  as  “vectors”.   Of course, one has to check that all the laws are satisfied.  This is routine: they are.  What plays the part of the zero vector is the zero constant function.
	
	Now, what is a subspace of a given vector space   V?

DEFINITION:  A subspace of the vector space  V  is  a  subset  W  of  V  such that  W,  together with the binary operations it inherits from  V ,  forms a vector space  (over the same field of scalars) in its own right.

EXAMPLES OF SUBSPACES:  
	(1)  Let  V  =  R3  and  W  =  {[x,y,x] | x, y  ˛ R}.
	(2)  Let  V  =  M(2,2)  and  W  =  {{{a,b},{0,a}} |  a , b  ˛ R }
	(3)  Let  V  =  C[0,1]  and  W  =  P_2  =  {f(x) = a  +  b x  + c x2 | a, b, c in  R}

In every case we have to verify that laws (1) - (10)  are valid for each subset  W  of  the v.s.  V.   This is unpleasant!!  Do we have to do all this work??   No,  and here is why:  

SUBSPACE  TEST:   Let  W  be a nonempty subset of the v.s.  V.   Then  W  is a  subspace of V  if and only if 
	(1)  For all u, v  ˛ W,  u + v ˛ W  (this is really closure of addition on elements of W).
	(2)  For all u  ˛ W  and scalars  a,  a*u  ˛ W  (this  is really closure of scalar multiplication on elements of W).
	
	It isn't too hard to see what the Subspace Test works:  certainly, any subspace  W of  V  must satisfy these conditions since it is a vector space in its own right with the operations inherited from  V.  Conversely,  if the nonempty subset W of  V  satisfies these conditions,  then vector space properties  (1) and (6) follow.  Properties  (2)-(3)  and (7)-(10)  follow simply because elements of W are elements of V  and they hold for elements of V.  The only properties to be checked are (4) and (5).   Since  W  is nonempty,  there is an element v ˛ W.  Now one of the Simple Facts above is that (-1)v = -v,  the additive inverse of v as an element of V.  By condition (2) of the Subspace Test, (-1)v ˛ W.  Therefore -v  ˛  W.   By condition (1) of the Subspace Test,  v + (-v)  =  0 (the  zero vector of the vector space  V)  also belongs to W.  It follows that  W  contains an additive identity, namely, the very same one as  V has, and that every element of  W  has  an additive inverse in W.  This disposes of vector space laws (4) and (5)  and shows that W is a vector space in its own right,  hence a subspace of V.
	
APPLICATION:  A very common occurrence is a linear combination (l.c.) of vectors  v1, v2,  ... , vn,  by which we mean an expression of the form,  for suitable scalars a1, a2, ...an,
	
		a1 v1 + a2 v2 + ... + an vn.
		
The set of all possible linear combinations of the vectors v1, v2, ... ,vn  has a name:  the linear span of v1, v2, ..., vn.

DEFINITION:  The linear span of vectors v1, v2, ...., vn  is the set of all vectors

	span{v1, v2, ..., vn}  = { a1 v1 + a2 v2 + ... + an vn | a1, a2,  ..., an  are scalars}.
	
Note that the span depends on the field of scalars in use.  Here is the application we referred to:

FACT:  If  v1, v2, ..., vn are vector in the v.s.  V  then  lin{v1, v2, ..., vn} is a subspace of V.

This is a fact that you should be able to prove.  It is also in your text.

ANOTHER IDEA:  if V is a vector space such that  V = lin{v1, v2, ..., vn},  then the vectors v1, v2, ..., vn  are said to SPAN  the vector space V.

EXAMPLE:   {1,0,0},  {0,1,0},  and {0,0,1}   span the vector space  R3.   As a matter of fact,  so does the set of vectors  {1,0,0},  {0, 1,0},{0,0,1}, {1,1,2}.   Why do you think that one would prefer the first spanning set?

DEFINITION:  A set of vectors v1, v2, ... , vn   is a BASIS  of the vector space  V  if this set spans  V  and  no subset of  v1, v2, ... vn  spans  V.

Subspaces and Matrices

Suppose we are given a matrix A.  For the sake of definiteness, we will work with the following matrix A:

A={{1,-2,0,0,3},{2,-5,-3,-2,6},{0,5,15,10,0},{2,6,18,8,6}}

Here are three very important vector spaces that are associated with the matrix  A.   Let us say that   A =  [C1, C2, ..., Cn]  where the Ai's  are the columns of A, and that  
A = [R1, R2, ..., Rm]T  where the Ri's  are the rows of  A.  Then:

	The column space of A:   C(A) =  lin{C1, C2, ..., Cn}
	
	The row space of A:  R(A) = lin{R1, R2, ..., Rn}
	
	The null space of A:  N(A) = { v ∈ Rn   A.v = 0}  
(Note: replace R by C in the case of complex scalars in all that follows.  We have to use one symbol, so we default to R.)
	
R(A) and C(A) are definitely vector spaces.  As a matter of fact, R(A) is a subspace of  Rm  and C(A)  is a subspace of Rn. (How come?) 

EXAMPLE:  Describe the row space, column space and null space of the simple matrix of the following cell in simple geometrical terms.  Where do these subspaces live?  For the null space, by the way, it might help to click on the cells after the definition of

B = {1, 3, 1}

v = {x,y,z}

B.v

Now let's take a look at a harder example, namely, the matrix  A  we defined above.  How would you define the row space and column space of  A?  Write this out on a separate piece of paper.  It might help to use the following comman

MatrixForm[A]

NullSpace[A]

Now for another idea about spanning sets.  Remember the examples  of the last section (look them up).  There was something undesirable about one spanning set:  it had unnecessarily many vectors in the list.  A spanning set which does not have too many vectors  is called a basis of the vector space it spans.  Thus:

DEFINITION:  A basis for the vector space  V is  a  minimal spanning set, that is,  a spanning set which has the property that no smaller set will span the space.

Give an example of a basis for R3.  Now  for a key question:  how can we spot redundancies in the list of rows or columns of a given matrix like our A?  I  claim the answer lies in the following cell:

RowReduce[A]

Exercises

Find the row space, the column space, and the null space of the following matrices.  Are there any relationships between the dimensions of (the number of independent vectors in) these spaces ?

1   0   2     7    5
1)  A =     3   3   -1   -3   0
               2   1   -2    0    5

3   2    10   5
2)  B =    1   1     0    2
               4   -2   8    0
               -3  5    -2   5

 2     2    1    0
              1     0    2    1
3)  C =   3    -1   -1    0
              -2    0    0    2
              4    -4    2    1