Coordinate Systems S F Ellermeyer July 10, 009 These notes closely follow the presentation of the material given in David C Lay s textbook Linear Algebra and its Applications (rd edition) These notes are intended primarily for in-class presentation and should not be regarded as a substitute for thoroughly reading the textbook itself and working through the exercises therein Coordinate Systems Recall that a basis for a vector space, V, is a set of vectors in V that is linearly independent and spans V Theorem 1 (The Unique Representation Theorem) If V is a vector space and B = fv 1 ; v ; ; v n g is a basis for V, then each vector in V can be written as a linear combination of the vectors in B in a unique way In other words, if v is any vector in V, then there exist unique scalars c 1 ; c ; : : : ; c n such that v =c 1 v 1 + c v + + c n v n Proof Since B spans V, we know that every vector in V is a linear combination of the vectors in B We need to show that there is only one possible way to write this linear combination (for each vector in V ) To this end, suppose that v is a vector in V and suppose that v =c 1 v 1 + c v + + c n v n and v =d 1 v 1 + d v + + d n v n 1
Then, since v v = 0, we have (c 1 d 1 ) v 1 + (c d ) v + + (c n d n ) v n = 0 However, since the set B is linearly independent, we can then conclude that c 1 d 1 = c d = = c n d n = 0 This means that c 1 = d 1, c = d,, c n = d n Thus, there is only one way to write v as a linear combination of the vectors in B De nition Suppose that B = fv 1 ; v ; ; v n g is a basis for the vector space V and suppose that x is a vector in V The unique set of scalars c 1 ; c ; : : : ; c n such that x =c 1 v 1 + c v + + c n v n are called the coordinates of the vector x with respect to the basis B, and the vector c 1 c [x] B = 6 7 5 V n c n is called the coordinate vector of x relative to the basis B
Example Let b 1 = 1 8 and b = Then B = fb 1 ; b g is a basis for V Find the coordinate vector of the vector x = 8 relative to the basis B 0
Example Let e 1 = 1 0 and e = Then E = fe 1 ; e g is a basis for V (called the standard basis for V ) Find the coordinate vector of the vector x = 8 relative to the basis E 0 1
A Graphical Interpretation of Coordinates The standard basis, E = fe 1 ; e g, for V corresponds to standard graph paper for < The picture below shows a graphical interpretation of the vector x = 8 being expressed as a linear combination of e 1 and e 5
The basis, B = fb 1 ; b g, for V (where b 1 and b are the vectors in the above example) corresponds to non standard graph paper for < The picture below shows a graphical interpretation of the vector x = 8 being expressed as a linear combination of b 1 and b 6
Changing Coordinates in V n Via Matrix Multiplication Let us take another look at the example that we have been studying: We are considering the vector x = V 8 We have seen that if B is the basis B= fb 1 ; b g = 1 8 0 ; then the coordinate vector of x relative to B is [x] B = In other words, We can write the above equation as 1 0 x = 8 If we de ne then we can simply write x =b 1 + b, = b 1 b [x]b P B = b 1 b, x =P B [x] B Since the columns of P B are linearly independent, then we know that P B is an invertible matrix We thus conclude that P 1 B x = [x] B This leads us to the following theorem Theorem 5 Suppose that B = fb 1 ; b ; : : : ; b n g is a basis for V n and de ne Then, for any vector x V n, we have P B = b 1 b b n and x =P B [x] B P 1 B x = [x] B 7
Remark 6 For the have the standard basis E = fe 1 ; e ; : : : ; e n g, we have P E = e 1 e e n = In Therefore, for any vector x V n, we have x =P E [x] E = I n [x] E = [x] E 8
Example 7 Let B = fb 1 ; b ; b g be the basis for V consisting of the vectors 1 0 b 1 = 5, b = 0 5, and b = 5 0 6 1 Find the coordinate vector of the vector 5 x = 7 5 6 relative to the basis B Solution 8 De ning we observe that P B = 1 0 0 0 6 1 5, [x] B = P 1 B x 0 = 1 1 1 6 6 1 0 1 = 1 5 To verify that this is correct, we observe that 5 5 7 6 5 x = b 1 + 1 b + b 9
The Coordinate Mapping If B = fb 1 ; b ; : : : ; b n g is a basis for a vector space V, then the mapping from V into V n de ned by x 7! [x] B is called the coordinate mapping with respect to the basis B This mapping is a linear transformation that is one to one and onto V n (as will be proved below) A one to one linear transformation from a vector space V onto a vector space W is called an isomorphism of V and W Thus, the coordinate mapping x 7! [x] B is an isomorphism of V and V n Theorem 9 If V is a vector space with basis B = fb 1 ; b ; : : : ; b n g (consisting of exactly n vectors), then the coordinate mapping x 7! [x] B is an isomorphism of V and V n Proof Let T : V! V n be de ned by T (x) = [x] B for all x V To prove that T is an isomorphism of V and V n, we must show that T is a linear transformation that is one to one and onto V n First we prove that T is a linear transformation: Suppose that x and y are two vectors in V with x =c 1 b 1 + c b + + c n b n and y =d 1 b 1 + d b + + d n b n Then x + y = (c 1 + d 1 ) b 1 + (c + d ) b + + (c n + d n ) b n, and for any scalar k, kx =kc 1 b 1 + kc b + + kc n b n 10
Thus, and T (kx) = [kx] B = 6 T (x + y) = [x + y] B c 1 + d 1 c + d = 6 7 5 c n + d n c 1 c 7 6 kc 1 kc kc n = 6 c n 7 5 + 6 d 1 d d n = [x] B + [y] B = T (x) + T (y) 7 5 = k 6 c 1 c c n 7 5 7 5 = k [x] B = kt (x) This shows that T is a linear transformation To show that T is one to one, suppose that x and y are vectors in V such that T (x) = T (y) Then [x] B = [y] B This immediately implies that x = y, thus showing that T is one to one To show that T is onto V n, we must show that every vector in V n is the image (under T ) of at least one vector in V To this end, let be a vector in V n Then let 6 c 1 c c n 7 5 x =c 1 b 1 + c b + + c n b n 11
It is clear that x V and that T (x) = 6 This shows that T is onto V n We conclude that V and V n are isomorphic to each other and that the coordinate mapping x 7! [x] B is an isomorphism of V and V n c 1 c c n 7 5 1
Example 10 Recall that P = P (<) is the vector space consisting of all polynomial functions with degree less than or equal to (and with domain <) Thus, every function p P is de ned by a formula of the form p (t) = a 0 + a 1 t + a t It is easily observed that the set of three functions E= fp 0 ; p 1 ; p g de ned by p 0 (t) = 1 p 1 (t) = t p (t) = t is a basis for P E is called the standard basis for P The coordinate vector of an arbitrary function p P (as written above) relative to the basis E is a 0 [p] E = a 1 5 a Thus, P is isomorphic to V 1
Example 11 Let B= fq 0 ; q 1 ; q g be the set of functions in P de ned by q 0 (t) = 1 q 1 (t) = t 1 q (t) = (t 1) Show that B is a basis for P and nd the coordinate vector of an arbitrary function p P relative to the basis B Solution 1 Suppose that p P is the function de ned by p (t) = a 0 + a 1 t + a t We want to show that there exist unique scalars b 0, b 1, and b such that p can be written as This is equivalent to Thus we must have p (t) = b 0 + b 1 (t 1) + b (t 1) p (t) = (b 0 b 1 + b ) + (b 1 b ) t + b t b 0 b 1 + b = a 0 b 1 b = a 1 b = a The augmented matrix for the above linear system is 1 1 1 a 0 1 1 1 a 0 0 1 a 1 5 ~ 0 1 0 a 1 + a 0 0 1 a 0 0 1 a ~ 1 1 0 a 0 a 0 1 0 a 1 + a 0 0 1 a 5 ~ 5 1 0 0 a 0 + a 1 + a 0 1 0 a 1 + a 0 0 1 a The fact that the rst three columns of the augmented matrix are equivalent to the identity matrix shows that B is a basis for P We have also discovered that a 0 + a 1 + a [p] B = a 1 + a a 5 1 5
As a concrete example, suppose that p P is the function de ned by p (t) = 5t 8t Then This means that [p] B = 9 1 8 5 p (t) = 9 1 (t 1) 8 (t 1) Remark 1 In Calculus, we learn that a function of the form p (t) = a 0 + a 1 t + a t can be written as a Taylor Series centered at 1: p (t) = p (1) + p 0 (1) (t 1) + p00 (1)! (t 1) This is the same thing that we have done in the preceding example Observe that if p (t) = a 0 + a 1 t + a t, then Thus p 0 (t) = a 1 + a t p 00 (t) = a p (1) = a 0 + a 1 + a p 0 (1) = a 1 + a p 00 (1) = a Therefore, the theory of Taylor Series tells us that p (t) = (a 0 + a 1 + a ) + (a 1 + a ) (t 1) + a! (t 1) 15
Problem 1 Consider the set of functions B= fp 0 ; p 1 ; p ; p g de ned by p 0 (t) = 1 p 1 (t) = t p (t) = (t ) The set of functions is a basis for P p (t) = (t ) 1 Use Linear Algebra to write the function p (t) = 9 + t 6t + t as a linear combination of the functions in the basis B Use your knowledge of Taylor Series to write the function p (t) = 9 + t 6t + t as a linear combination of the functions in the basis B 16