Coordinates and Basis

MA 2071

 

 

Purpose of these notes and assignment:

 

            illustrate the concepts and skills involved with coordinates and coordinate systems (also know as basis)

 

 

Definition:  given a vector space   V,  a basis (or coordinate system)  is a set of vectors  {v1, v2, v3, . . . vk} 

 with two properties:

 

            1) any vector  x  can be written as a unique combination of the vectors in V, or

 

                        x = c1v1 + c2v2+ . . . + ckvk                         equation (1)

 

            2)  the basis vectors are independent of one another. None can be written as a

                  combination of the others.

 

Remarks:  be careful to distinguish between vectors and scalars.  The vi s are vectors while the  cis are scalars (numbers). The latter are called  coordinates.

 

Example 1:  {  (1,1) ,  (1,-1)  }   is a basis for the vector space R2.  In the notation above,  v1 = (1 ,1)   and  v2 = (1, -1).  We know that it is a basis because we can satisfy the two properties of the definition as follows:

 

a)  any vector in R2  can be written as a combination of them.  Let  x = (x , y)  be an arbitrary vector in R2.  The equation

 

                                    X = (x,y) = c1v1 + c2v2  = c1(1 ,1)  +  c2 (1 , -1)

 

is really a system of two equations and two unknowns.  This is important to recognize.  This can be better seen by looking at it component by component:

 

                                    x = c1  + c2

                                    y = c1  -  c2

 

Does this system have a solution?? It is a nonhomogeneous system so the possibility exists that it doesn't.  Since x and y can literally be any numbers, we have to be ready for all possibilities!!!

We solve it the same way we have been solving systems throughout the course. Set up the augmented matrix and reduce it to Final Form.  Here, the system in matrix form looks like:

 

                                               

 

 

 

which means that the augmented matrix is

 

                                               

 

 

 

( you have to get used to the fact that x and y are treated as constants while the c's are the variables.)  Now you should be able to look at this system and see that it will always have a solution, no matter what x and y are.  Why? How do you end up in a "no solution" situation?  You would have to have, on the bottom row a situation like

 

 

 

 

where whatever is in the place of  &  is not 0.   Can this happen here? With one row operation (adding -1 times Row1 to Row2 ) we

arrive at

 

                                               

 

 

 

which should tell you it will have a unique solution. If it doesn't, we'll go another couple of  steps and get Final Form:

 

                                               

 

 

 

This Final Form shows that no matter what x and y are, we will have a unique solution. This means that any vector (x , y) can be written  as a combination of  (1 , 1) and  (1, -1). The first part of the definition has been satisfied.

 

All this can be used to summarize an Algorithm for Finding Coordinates of a vector relative to a basis:

 

            Problem:  find the coordinates of a vector  x relative to a basis  {v1, v2, v3, . . . vk} 

 

                                    step 1:  set up the augmented matrix    (v1 v2 v3 . . . vk | x)    This matrix has the basis vectors as columns

                                    on the left and the given vector x on the right

 

                                    step 2:  reduce it to Final Form (either with Gauss Jordan or  software)

 

                                    step 3:  Interpret the Final Form

 

                                               

 

 

 

                                    step 4: Read them off the matrix.

 

                                    Note:  if you had a row of 0's appear indicating either no solution or infinitely many, then either you made an arithmetic mistake or you did not really have a basis to begin with

 

            Notation for future reference: the coefficient portion with the basis vectors as columns will always be denoted by P in this course.

            That is,  

                                    P = (v1 v2 . . . vk)

 

          In matrix terms, we are solving the problem   Pc =  x    (key equation!)

 

Example 1:   find the coordinates of  ( 6,3,6) relative to  {  (1,0,-1) ,   (0,1,1),  (4,0,5)   }

 

                                    step 1:  set up the augmented matrix    (P | x)

 

                                                           

 

 

                                    step 2:  reduce it to Final Form

 

                                                           

 

 

                                    step 3 & 4 :  interpret the Final Form

 

the coordinates are   c1 = 2,  c2=3,  c3 = 1

 

            this means that  (6,3,6) = 2(1,0,-1)   +  3 (0,1,1)  +  1( 4,0,5)  which can be easily checked

 

 

 

Example 2:   find the coordinates of  ( 6,3,6) relative to  {  (1,0,-1) ,   (0,1,1),  (2,3,1)   }

 

                                    step 1:  set up the augmented matrix    (P | x)

 

 

                                                           

 

 

 

                                    step 2:  reduce it to Final Form

 

 

 

 

 

                                    step 3 & 4 :  the bottom row tells us we have no solution.  The vector  (6,3,6) cannot be built out of the three given vectors.

                                    There is actually more information here…. the third column in the Final Form is telling us that the vector (2,3,1) is really  2 times the first vector, (1,0,-1)  plus  3 times the vector  (0,1,1). ( the reader is urged to check this detail)  The three vectors given were not independent; one was a combination of the others.

 

                        For those who like a geometric view of things,  the three vectors , {  (1,0,-1) ,   (0,1,1),  (2,3,1)   } , all lie in the same plane.

                                    The vector  (6,3,6) does not lie in that plane.  There is no way to have vectors in the plane add up to a vector which is not in the plane!

 

Example 3:   find the coordinates of  ( 9,6,1) relative to  {  (2,1,-1) ,   (3,1,2),  (2,3,1), (9,6,1)   }

 

                                    step 1:  set up the augmented matrix    (P | x)

                                               

 

 

 

                                    step 2:  reduce it to Final Form

                                               

 

 

                                    step 3: interpret the Final Form

 

                                                It tells us the problem has a solution. It also tells us

                                    there is an arbitrary variable  (c4).  This is then telling us that      

                                    we don't need the fourth vector  (9, 6 , 1).  In fact the 4th column of the Final Form

                                    is telling us that the vector (9, 6 , 1)  is a combination of the first three, specifically it

                                    is 2 * (2 , 1, -1)  + 1*(3, 1, 2) + 1*(2, 3, 1).

 

                                                The upshot of all this is we don't have a basis; the vectors are not independent of one

                                    another.  (the fix to this would be to throw out the 4th vector – then we would have a vector).

 

 

 

Discussion:  How do we know if a set of vectors forms a basis for Rn??

 

                        none can be a combination of the others.  This means we want the rank of the P matrix to be the same as the number of vectors (k above)

 

                        we need to be able to write any vector in Rn as a combination of them which amounts to saying we must

                        always be able to solve the problem  Pc = x.  So we don't want to have a row of zeroes appear in the Final Form of P because we can't assume there will be a 0 in the reduced form of x, the right hand side.  This means the Final Form of P must indicate a rank of n for it.

 

                        In order that we have no arbitrary variables in the solution for the c's because if we do then the vectors in the claimed basis won't be independent – there will be repetition within them.  Thus there must be n columns in P – any more than that will cause arbitrary variables.

 

                        Putting all this together means that P must be nxn and it must have rank of n. So its Final Form has only one possibility, the Identity matrix,  In !  Anything else signifies that either there is no solution to some problems or the vectors are not independent.

 

Summary:  for a set  {v1, v2, v3, . . . vk}  to be a basis for Rn  we must have

 

        • k = n

 

        • the Final Form of  P = (v1 v2 v3 . . . vk) must be In 

 

any other result indicates we don't have a basis. There are either too many vectors, too few vectors or some vector is a combination of the others.

 

If you do have a basis then to find the coordinates of x relative to the basis, {v1, v2, v3, . . . vk}  ,

solve the problem

 

                        Pc = x                       (which symbolically is  c = P-1x  )

 

Its Final Form is guaranteed to   be  (In  | c )   The coordinates will be on the right hand side as in Example 1

 

Homework for Tuesday

 

page 272:   problems 1-3  (decide if a set is a basis or not),  problems 7 and 8  (finding  coordinates)

 

also:  a plane through the origin has a basis consisting of { (1, 0 , 4) ,  (0,1,9)  }    On the plane, I am at the location  (3,5).  Where am I in three dimensional space  (R3) ??