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 (The Unique Representation Theorem) If V is a vector space and B v,v 2,,v n 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,c 2,,c n such that v c v c 2 v 2 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 v c 2 v 2 c n v n and v d v d 2 v 2 d n v n. Then, since v v 0, we have c d v c 2 d 2v 2 c n d nv n 0. However, since the set B is linearly independent, we can then conclude that c d c 2 d 2 c n d n 0. This means that c d, c 2 d 2,, c n d n. Thus, there is only one way to write v as a linear combination of the vectors in B. Definition Suppose that B v,v 2,,v n is a basis for the vector space V and suppose that x is a vector in V. The unique set of scalars c,c 2,,c n such that x c v c 2 v 2 c n v n are called the coordinates of the vector x with respect to the basis B, and the vector c x B c 2 c n n
is called the coordinate vector of x relative to the basis B. 2
Example Let 0 b and b 2. 8 4 Then B b,b 2 is a basis for 2. Find the coordinate vector of the vector x 8 relative to the basis B.
Example Let 0 e and e 2. 0 Then E e,e 2 is a basis for 2 (called the standard basis for 2 ). Find the coordinate vector of the vector relative to the basis E. x 8 4
A Graphical Interpretation of Coordinates The standard basis, E e,e 2, for 2 corresponds to standard graph paper for 2. The picture below shows a graphical interpretation of the vector x 8 being expressed as a linear combination of e and e 2. 5
The basis, B b,b 2, for 2 (where b and b 2 are the vectors in the above example) corresponds to non standard graph paper for 2. The picture below shows a graphical interpretation of the vector x 8 being expressed as a linear combination of b and b 2. 6
Changing Coordinates in n Via Matrix Multiplication Let us take another look at the example that we have been studying: We are considering the vector We have seen that if B is the basis x 8 B b,b 2 8 then the coordinate vector of x relative to B is x B 4 2. In other words, x b 4b 2. We can write the above equation as If we define x 0 8 4 4 P B b b 2,,. 0 4 b b 2 x B. then we can simply write x P Bx 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 B x x B. This leads us to the following theorem., Theorem Suppose that B b,b 2,,b n is a basis for n and define P B b b 2 b n. Then, for any vector x n, we have x P Bx B and P B x x B. 7
Remark For the have the standard basis E e,e 2,,e n, we have P E e e 2 e n I n. Therefore, for any vector x n, we have x P Ex E I nx E x E. 8
Example Let B b,b 2,b be the basis for consisting of the vectors 0 2 b 2, b 2 0, and b. 0 6 Find the coordinate vector of the vector x 5 7 6 relative to the basis B. Solution Defining P B 0 2 2 0 0 6, we observe that x B P B x 2 0 6 6 2 0 5 7 6 2. To verify that this is correct, we observe that x b 2 b 2 b. 9
The Coordinate Mapping If B b,b 2,,b n is a basis for a vector space V, then the mapping from V into n defined by x 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 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 x B is an isomorphism of V and n. Theorem If V is a vector space with basis B b,b 2,,b n (consisting of exactly n vectors), then the coordinate mapping x x B is an isomorphism of V and n. Proof Let T : V n be defined by Tx x B for all x V. To prove that T is an isomorphism of V and n, we must show that T is a linear transformation that is one to one and onto n. First we prove that T is a linear transformation: Suppose that x and y are two vectors in V with x c b c 2 b 2 c n b n and y d b d 2 b 2 d n b n. Then x y c d b c 2 d 2b 2 c n d nb n, and for any scalar k, Thus, kx kc b kc 2 b 2 kc n b n Tx y x y B c d c 2 d 2 c n d n c d d 2 c 2 c n d n x B y B and Tx Ty 0
kc c Tkx kx B kc 2 k c 2 kx B ktx. kc n c n 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 Tx Ty. 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 n, we must show that every vector in n is the image (under T) of at least one vector in V. To this end, let be a vector in n. Then let x c b c 2 b 2 c n b n. It is clear that x V and that c c 2 c n c Tx c 2 c n. This shows that T is onto n. We conclude that V and n are isomorphic to each other and that the coordinate mapping x x B is an isomorphism of V and n.
Example Recall that P 2 P 2 is the vector space consisting of all polynomial functions with degree less than or equal to 2 (and with domain ). Thus, every function p P 2 is defined by a formula of the form pt a 0 a t a 2 t 2. It is easily observed that the set of three functions E p 0,p,p 2 defined by p 0t p t t p 2t t 2 is a basis for P 2. E is called the standard basis for P 2. The coordinate vector of an arbitrary function p P 2 (as written above) relative to the basis E is p E a 0 a. a 2 Thus, P 2 is isomorphic to. 2
Example Let B q 0,q,q 2 be the set of functions in P 2 defined by q 0t q t t q 2t t 2. Show that B is a basis for P 2 and find the coordinate vector of an arbitrary function p P 2 relative to the basis B. Solution Suppose that p P 2 is the function defined by pt a 0 a t a 2 t 2. We want to show that there exist unique scalars b 0,b, and b 2 such that p can be written as pt b 0 b t b 2t 2. This is equivalent to pt b 0 b b 2 b 2b 2t b 2 t 2. Thus we must have b 0 b b 2 a 0 b 2b 2 a b 2 a 2. The augmented matrix for the above linear system is a 0 0 2 a ~ 0 0 a 2 a 0 0 0 a 2a 2 0 0 a 2 ~ 0 a 0 a 2 0 0 a 2a 2 ~ 0 0 a 2 0 0 a 0 a a 2 0 0 a 2a 2. 0 0 a 2 The fact that the first three columns of the augmented matrix are equivalent to the identity matrix shows that B is a basis for P 2. We have also discovered that p B a 0 a a 2 a 2a 2 a 2. As a concrete example, suppose that p P 2 is the function defined by pt 4 5t 8t 2. Then
This means that p B 9 2 8. pt 9 2t 8t 2. Remark In Calculus, we learn that a function of the form pt a 0 a t a 2 t 2 can be written as a Taylor Series centered at : pt p p t p t 2. 2! This is the same thing that we have done in the preceding example. Observe that if pt a 0 a t a 2 t 2, then p t a 2a 2 t p t 2a 2. Thus p a 0 a a 2 p a 2a 2 p 2a 2. Therefore, the theory of Taylor Series tells us that pt a 0 a a 2 a 2a 2t 2a 2 2! t 2. Problem Consider the set of functions B p 0,p,p 2,p defined by p 0t p t t 4 p 2t t 4 2 p t t 4. The set of functions is a basis for P.. Use Linear Algebra to write the function pt 9 2t 6t 2 2t as a linear combination of the functions in the basis B. 2. Use your knowledge of Taylor Series to write the function pt 9 2t 6t 2 2t as a linear combination of the functions in the basis B. 4