Overview Last time we defined a basis of a vector space H: Definition The set v , , v is a basis for H if { 1 ··· p} v , , v is linearly independent, and { 1 ··· p} Span v , , v = H { 1 ··· p} We recalled algorithms (§2.8, §4.3) to find a basis for the null space and the column space of a matrix, and we stated the Unique Representation Theorem: Given a basis for H, every vector in H can be a written as a linear combination of basis vectors in a unique way. The coefficients of this expression are the coordinates of the vector with respect to the basis. Question Given bases and for H, how are [x] and [x] related? B C B C Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 1 / 29 (Lay, §4.4, §4.7) Coordinates Theorem (The Unique Representation Theorem) Suppose that = v ,..., v is a basis for a vector space V . Then each B { 1 n} x V has a unique expansion ∈ x = c v + c v (1) 1 1 ··· n n where c1,..., cn are in R. We say that the c are the coordinates of x relative to the basis , and we i B c1 write [x] = . B . cn Coordinates give instructions for writing a given vector as a linear combination of basis vectors. Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 2 / 29 Different bases determine different coordinates... 1 1 1 0 Suppose = , , and as always, = , . B {"0# "2# } E {"0# "1# } E E E E x x b2 e2 e1 b1 Standard graph B-graph paper 2 1 1 4 If [x] = , then x = 2b1 + 2b2 = 2 + 2 = B "2# "0# "2# "4# E E E 4 1 0 4 Similarly, [x] = , so x = 4e1 + 4e2 = 4 + 4 = E "4# "0# "1# "4# E E E Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 3 / 29 ...but some things stay the same Even though we use different coordinates to describe the same point with respect to different bases, the structures we see in the vector space are independent of the chosen coordinates. Definition A one-to-one and onto linear transformation between vector spaces is an isomorphism. If there is an isomorphism T : V V , we say that V and 1 → 2 1 V2 are isomorphic. Informally, we say that the vector space V is isomorphic to W if every vector space calculation in V is accurately reproduced in W , and vice versa. For example, the property of a set of vectors being linearly independent doesn’t depend on what coordinates they’re written in. Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 4 / 29 Isomorphism Theorem Let = b , b ,..., b be a basis for a vector space V . Then the B { 1 2 n} coordinate mapping P : V Rn defined by P(x) = [x] is an → B isomorphism. What does this theorem mean? V and Rn are both vector spaces, and we’re defining a specific map that takes vectors in V to vectors in Rn. This map ...is a linear transformation ...is one-to-one (i.e., if P(u) = 0, then u = 0) ...is onto (for every v Rn, there’s some u V with P(u) = v) ∈ ∈ Every vector space with an n-element basis is isomorphic to Rn. Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 5 / 29 Very Important Consequences If = b ,..., b is a basis for a vector space V then B { 1 n} A set of vectors u1, , up in V spans V if and only if the set of { ··· } n the coordinate vectors [u1] ,..., [up] spans R ; { B B} A set of vectors u , , u in V is linearly independent in V if and { 1 ··· p} only if the set of the coordinate vectors [u1] ,..., [up] is linearly { B B} independent in Rn. An indexed set of vectors u , , u in V is a basis for V if and { 1 ··· p} only if the set of the coordinate vectors [u1] ,..., [up] is a basis { B B} for Rn. Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 6 / 29 Theorem If a vector space V has a basis = b ,..., b , then any set in V B { 1 n} containing more than n vectors is linearly dependent. Theorem If a vector space V has a basis consisting of n vectors, then every basis of V must consist of exactly n vectors. That is, every basis for V has the same number of elements. This number is called the dimension of V and we’ll study it more tomorrow. Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 7 / 29 Changing Coordinates (Lay §4.7) When a basis is chosen for V , the associated coordinate mapping onto B Rn defines a coordinate system for V . Each x V is identified uniquely by ∈ its coordinate vector [x] . B In some applications, a problem is initially described by using a basis , B but by choosing a different basis , the problem can be greatly simplified C and easily solved. We want to study the relationship between [x] , [x] in Rn and the vector B C x in V . We’ll try to solve this problem in 2 different ways. Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 8 / 29 Changing from to coordinates: Approach #1 B C Example 1 Let = b , b and = c , c be bases for a vector space V , and B { 1 2} C { 1 2} suppose that b = c + 4c and b = 5c 3c . (2) 1 − 1 2 2 1 − 2 2 Further, suppose that [x] = for some vector x in V . What is [x] ? B "3# C Let’s try to solve this from the definitions of the objects: 2 Since [x] = we have B "3# x = 2b1 + 3b2. (3) Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 9 / 29 The coordinate mapping determined by is a linear transformation, so we C can apply it to equation (3): [x] = [2b1 + 3b2] C C = 2[b1] + 3[b2] C C We can write this vector equation as a matrix equation: 2 [x] = [b1] [b2] . (4) C C C "3# h i th Here the vector [bi ] becomes the i column of the matrix. C This formula gives us [x] once we know the columns of the matrix. But C from equation (2) we get 1 5 [b1] = − and [b2] = C 4 C 3 " # "− # Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 10 / 29 So the solution is 1 5 2 13 [x] = − = or C 4 3 3 1 " − #" # "− # [x] = P [x] C C←B B 1 5 where P = − is called the change of coordinate matrix from C←B 4 3 " − # basis to . B C Note that from equation (4), we have P = [b1] [b2] C←B C C h i Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 11 / 29 The argument used to derive the formula (4) can be generalised to give the following result. Theorem (2) Let = b ,..., b and = c ,..., c be bases for a vector space V . B { 1 n} C { 1 n} Then there is a unique n n matrix P such that × C←B [x] = P [x] . (5) C C←B B The columns of P are the -coordinate vectors of the vectors in the C←B C basis . That is B P = [b1] [b2] [bn] . (6) C←B C C ··· C h i Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 12 / 29 The matrix P in Theorem 12 is called the change of coordinate matrix from to C←B. B C Multiplication by P converts -coordinates into -coordinates. C←B B C Of course, [x] = P [x] , B B←C C so that [x] = P P [x] , B B←C C←B B whence P and P are inverses of each other. B←C C←B Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 13 / 29 Summary of Approach #1 The columns of P are the -coordinate vectors of the vectors C←B C in the basis . B Why is this true, and what’s a good way to remember this? Suppose = b ,..., b and = c ,..., c are bases for a vector B { 1 n} C { 1 n} space V . What is [b1] ? B 1 0 [b ] = . 1 . B . 0 We have [b1] = P [b1] , C C←B B so the first column of P needs to be the vector for b1 in coordinates. C←B C Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 14 / 29 Example Example 2 Find the change of coordinates matrices P and P for the bases C←B B←C = 1, x, x 2 and = 1 + x, x + x 2, 1 + x 2 B { } C { } of P2. Notice that it’s “easy" to write a vector in in coordinates. C B 1 0 1 [1 + x] = 1 , [x + x 2] = 1 , [1 + x 2] = 0 . B B B 0 1 1 Thus, 1 0 1 P = 1 1 0 . B←C 0 1 1 Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 15 / 29 Example 3 (continued) Find the change of choordinates matrices P and P for the bases C←B B←C = 1, x, x 2 and = 1 + x, x + x 2, 1 + x 2 B { } C { } of P2. Since we just showed 1 0 1 P = 1 1 0 , B←C 0 1 1 we have 1/2 1/2 1/2 1 − − P = P = 1/2 1/2 1/2 . C←B B←C − 1/2 1/2 1/2 − Dr Scott Morrison (ANU) MATH1014 Notes Second Semester 2015 16 / 29 Suppose now that we have a polynomial p(x) = 1 + 2x 3x 2 and we want − to find its coordinates relative to the basis.