Vector Spaces 

In this chapter we will look for some underlying properties of the columns and row of a matrix that models a physical process to determine the properties of solutions.  There will be very little new when it comes to solving a system of equations.  We are going to use what we learned from Chapter 1 to further study and refine our analysis of problems.   

Consider the following water tower system.    Here .  and will denote the amount of water in tube and respectively at hour k. will denote the amount of water going from  

to during hour k and is the amount of water flowing from tank to during this time. 

Also,  flows into tank  and another flows into tank .   We would like to set up a set of equation to determine the level of each tank at hour k+1. 

      =  

=
We would like to know what values the difference between hour k+1 and hour k can take. 

Set the components equal to a, b and c respectively  

>	with(linalg):

 

>	A:=matrix([[-1,1,0],[1/2,-1,1],[1/2,0,-1]]);

 

>	B:=matrix([[a],[b],[c]]);

 

>	x:=augment(A,B);

 

>	gausselim(x);

 

So we know this requires that c=-a-b or only differences of the from  

.  We need to study further the columns of the coefficient matrix to determine how we would know what is going on and why.   

Definition 2.2 A vector space is a set of objects V along with two operations, addition and scalar multiplication, that satisfy the following.  Let u, v and w be any elements in V and let a and b be arbitrary scalars (real or complex numbers).  Then  

i) u + v is in V (V is closed under addition). 

ii) au is in V (V  is closed under scalar multiplication  

iii) u+v=v+u 

iv)(u+v) + w = u + (v + w). 

v) There is a unique element 0 in V such that u + 0 = 0 + u = u. 

vi) There is a unique element -u = (-1)u in V such that u+(-u) = (-u) + u = 0. 

vii) (ab)u= a(bu). 

viii) (a + b)u = au + bu. 

ix)  a(u + v) = au + av 

x) (1)u = u 

This will be called a space for short.   

Do you know of any vector spaces that we have studied so far in this class? 

One is obvious and that is We studied these vector spaces in the first chapter.  The second one that we have studied is the set of all or matrices.  As stated in Chapter 1 these set of matrices satisfy the 10 properties given in Definition 2.2.  We will be interested in subset of these vector spaces that are vector spaces in their won right.   

Definition 2.3  Any subset of elements of a vector space V that is closed with respect to forming linear combinations (closed with respect to addition and scalar multiplication) is called a subspace 

of V.  A subspace is a vector space in its own right.   

In and can you think of any subsets that form subspaces?   Look on Page 216 for subsets that are subspaces of .  On page 221 of your text we find a discussion of some subsets that are subspaces.  In general you should think of a point, lines, and planes with a special property.  What property is this?  Look on page 219 to see what property every subspace must have.  This property is discussed at the bottem of page 219.  Having this property does not make it a subspace but without it you have no chance of the subset being a subspace.    

Consider the following sets  

 where a and b are any real numbers. 

We will see shortly that this subset is a subspace.  We need to look at the pattern of this set of elements.  The first entry a is arbitary and the last element b is arbitrary. 

The second entry is then the first entry plus 2 times the last.  If we let a=2 and b =3 then the second entry must be 2 + 2*3 = 8.   or The letters a anb b above are not important the important thing is the pattern.  The following is another vector in this set Consider now a different set of numbers in three space that satistifes the following pattern.  

 where a and b are any real numbers. 

On the other hand this subset is not a subspace.  Here a and b can be any numbers and to be in this set a vector in three space must be one more than a , b is arbitrary and the third enty is the arbitrary a plus b.  If a=3 and b==1 then a vector in this set can be given by Again remember the letters a and b where selected to illustrate the pattern.  Another vector would be  If you want two arbitrary elements in the set we need only come up with different letters that follow the pattern.   

Lets now look at 2x2 matrices 

 where a, b and c are any real numbers 

This subset can be shown to be a subspace.    We should remember that the above set is the set of all 2x2 symmetric matrices.    A numeric example can be found by letter a =3 , b=1/4 and c=5.    Another arbitrary element of the set can be found by using the letters d,e and f in the following way..   Consider now another set of 2x2 matrices.   

where a and b are any real numbers. 

As above this fourth set is not a subspace.    This set is not all symmetric matrices since the element in the first row and first column must be one more than an arbitrary elment a.    Here if a=2, b=4 and c=3.    and another vecctor in this space would be  

Now lets take it a little further .   

Consider the subset of the set of all nxn matrices that consists of all skew-symmetric matrices.  Here we know that is the correct representation of the element in the set.   

Is this subset a subspace? 

New consider the set of all nxn matrices that are non-singular.  Is this subset a subspace?  You should remember that a matrix in this set is such that det(A) is not zero is the best way of things of the set of non-singular matrices.  

There was no explination as to why certain sets were subsets and others were not susets.    We are about to take a closer look at how to determine when we have a subspace and when we do not have a subspace.   We will find there are two ways of determining when many sets are subspaces.  In other cases was can use only one of the techniques.    

Definition 2.5  For a given set of m vectors {,...}, in a vector space, the set S of all possible linear combinations of the vectors is call the span of the set.   

z= +...+ where the c's are real of complex numbers. 

The set {,...,} is called a spanning set for S. 

The span for a set of vectors is a subspace of a vector space.   

= span{   }  

u =
z =
then   =
So this is back in the span.  This will help us to determine if a subset is subspace.  We can thus show a subset is a subspace if it can be show that a linear combination of vectors is again in the set as we just did.  One we show that the span of a set of vectors is always a subspace we have arrived at a second way of showing that a subset is a subspace.  i.e. if we can show that an element of a subset can be written as an arbitrary linear combination of vectors in the space then it is a subspace.   

Return to the vectors of the form 

=  
 

In terms of span then we know this subset is a subspace because it is in the span of the vectors 

and
 

Let us know look at the vectors of the from 

= + + =
What is the spanning set?    

We will not look at determining whether the above sets form a vector space.  Let us look at the first set.   When possible we will use illustrate both methods.   

>	with(LinearAlgebra):

 

>	v1:=<a,a+2*b,b>;

 

(1.1)

 

>	v2:=<c,c+2*d,d>;

 

(1.2)

 

v1 and v2 thus are two arbitrary elements of this set.  The following is checking by using an arbitrary linear combination of the vectors v1 and v2. 

>	v3:=alpha*v1 +beta*v2;

 

(1.3)

 

>

 

>

 

Notice if you expand out the second term of v3 , we  can show that the linear combination is of the same form as the origional vectors v1 and v2.  i.e. the second element is the first plus 2 time sthe last.  It takes a little of look at the second entry but it is really there.   

You can see the pattern a little better but the work will be a little longer if we first look at the sum and then look at a scalar times an element of the set. 

>

v1+v2;

 

(1.4)

 

You can see that the middle term the sum of the first and 2 times the last term easier here than above.  

Now we look at a scalar multiple.  

>

alpha*v1;

 

(1.5)

 

By multiplying through by the alpha we can see that the pattern is correct.  i.e. the middle term is the first term plus 2 times the last term.  This tells is that it is a subset of the vector space.   

Now we will look at the second subset. 

>	x:=<a+1,b,a+b>;

 

(1.6)

 

>	y:=<c+1,d,c+d>;

 

(1.7)

 

>

x+y;

 

(1.8)

 

We can see that the first term is not the same from since the first element has a +c+2 not a+c +1.  So this is not of the correct form and hence the set is not a subspace but just a subspace.   Another way to see it is not a subsapce is that the zero vector is not a member of this set.  The zero vector not being in the set is enough to tell you that it is not a subpace but zero being in the set is not enough to tell you that it is a subpace.   

We now will look at the other two examples.   Here I will again give the short way first and the longer way second.  After this I will not do the longer way.   

>	A:=<<a,b>|<b,c>>;

 

(1.9)

 

>	B:=<<e,f>|<f,g>>;

 

(1.10)

 

>	C:=alpha*A+beta*B;

 

(1.11)

 

As you can see the form of the matrices of this subset have the same elements in the off diagonal elements while the diagonal elements are arbitrary.    So the elements in the subset are closed under linear combinations which tell us it is a subspace.   

>

A+B;

 

(1.12)

 

>

alpha*A;

 

(1.13)

 

Again you can see that the sum and the scalar product are both 2x2 symmetric matrices.  So we have a subspace.   

>

 

>	E:=<<a+1,b>|<b,c>>;

 

(1.14)

 

>	F:=<<d+1,e>|<e,f>>;

 

(1.15)

 

both E and F are of the same form so in the same subset.  Now does this subset form a subspace. 

>

G:=E+F;

 

(1.16)

 

As you can see the sum of elements is not of the same form since the element in the 1,1 position has a 2 not a 1 in this position.  Hence, this subset is not a subspace.  

Again, the zero 2x2 matrix is not a member of this set and this is enough to tell you that the subset is not a subspace.   

Above, you were ask about the set of skew-symmetric matrices and non-singular matrices. 

There is not easy way to us Maple commands but we can use the text commands to look at skew-symmetric matrices.  The non-singular matrices will be addressed by using Maple commands. 

1.  A is skew symmetric i.e.   = -A. 

2.  B is skew symmetric  i.e.   = -B.  

Now consider a linear combination of A and B.  = + β=-= -+)   which is skew-symmetric.  So the set of skew-symmtric is a subspace. If you have a problem with the linear combination then look at the sum  and a scalar multiple separately.    

Finally, we will look at the subset of all non-singular matrices.  First recall that the set of all non-singular matrices will have a non-zero determinate.   We will look at the set of non-singular 2x2 matrices. 

>	A:=IdentityMatrix(2);

 

(1.17)

 

>

B:=-1*A;

 

(1.18)

 

>	Determinant(A);

 

(1.19)

 

>	Determinant(B);

 

(1.20)

 

>

 

We see that both matrices are non-singular but what about the sum of the two. 

>

C:=A+B;

 

(1.21)

 

>	Determinant(C);

 

(1.22)

 

>

 

So the sum is a singular matrix so the subset is not a subspace.    To show things are not a subspace then you can go to an example here it is not ture.  If you are looking at the set of nxn matrices showing it is true for 2x2 or 3x3 matrices is not enough to show it holds in general.   

You can thus show that a set of vectors form a subspace by finding a spanning set or by showing that it is closed under addition and scalar multiplication. 

By being closed under addition and scalar multiplication we mean that the pattern of the vector does not change when add two vectors form the set or if we multiply a vector in the set by an arbitrary constant.  This is what we have just done.   

I will now use the spanning sets to show that the  first sets are subspaces.  I did this by using text but will go over it again using Maple commands.  I find that it is shorter to find a spanning set for these two examples that to show they are closed under addition and scalar multiplication.  Which way you do the problems is up to you.  The set of skew-symmetric matrices must be done by showing the set is closed under addition and scalar multiplication.  

>	v1=a*<1,1,0>+b*<0,2,1>;

 

(1.23)

 

The spanning set is then  

>	<1,1,0>; <0,2,1>;

 

 

	(1.24)

 

For the set of all 2x2 matrices we have  

>	A:=<<a,b>|<b,c>>;

 

(1.25)

 

>	A=a*<<1,0>|<0,0> >+ b*<<0,1>|<1,0>>+ c*<<0,0>|<0,1>>;

 

(1.26)

 

The spanning set will be the three matrices 

>	<<1,0>|<0,0>>; <<0,1>|<1,0>>; <<0,0>|<0,1>>;

 

 

 

	(1.27)

 

>

 

Finding a spanning set will show that a subset of elements in a vector space from a subspace.   

We will be looking a different susets to see if they are subspace for the remainder of the course. 

A closer look at the model which was examined at the beginning of this lecture reveals that the subset of all vectors in the difference of day k+1 and day k form a subspace.  This can be verified by look at linear combinations or by finding a set of spanning vectors.   

We will also look at what more we can find from the model.  First look at the matrix  

>

 

>	A:=<<-1,1/2,1/2>|<1,-1,0>|<0,1,-1>>;

 

(1.28)

 

>	ca1:=Column(A,1);

 

(1.29)

 

>	ca2:=Column(A,2);

 

>	ca3:=Column(A,3);

 

 

	(1.30)

 

A moments thought will tell you that the difference between day k+1 and day k is in the span of the three columns span(ca1, ca2, ca3) 

Also, every difference will be of the form. 

>	dif:=<<a,b,-a-b>>;

 

(1.31)

 

We can write this vector as a linear combination of the form 

>	dif:=a*<1,0,-1>+b*<0,1,-1>;

 

(1.32)

 

We are going to the first vector in the above linear combination a location in Maple 

>	ca4:=<1,0,-1>;

 

(1.33)

 

The question is span(ca1,ca2,ca3) =span(ca3,ca4).  Also, how can we prove that the span of the three vectors is the same subspace as the two vectors.  One way of doing this is to prove that ca1 and ca2 can be written as a linear combination a ca3 and ca4.  i.e. ca3 + ca4 = ca1 and ca3 +ca4 = ca2.  Another may to consider this is to say that a system has a solution.  Lets see if we can find and .  Next, we need to find and . 

>	sys1:=<ca3|ca4|ca1>;

 

(1.34)

 

>	sol:=LinearSolve(sys1);

 

(1.35)

 

>	1/2*ca3+(-1)*ca4;

 

(1.36)

 

>	sys2:=<ca3|ca4|ca2>;

 

(1.37)

 

>	sol2:=LinearSolve(sys2);

 

(1.38)

 

>	-1*ca3+ca4;

 

(1.39)

 

i.e. =1/2and =-1 while = -1 and =1.  So the span of the three vectors is the same as the span of the two vectors.    In the next section we will introduce some more definitions to better understand why we like to use the two vectors over the three vectors.   

>

 

Exercises: 

Exercise 2.1.1(a),(b),(c),(d),(g), 2.1.4, 2.1.5, and 2.1.11.  The instructions are as usual for turning in the assignment.  The assignment is due on October 13,   2008.  

>