There's a transition matrix for that. Let the β-basis vector bj have standard coordinates bj = (b1j; b2j; : : : ; bnj), so 2 3 b1j 6b2j 7 [bj] = 6 . 7 Change of coordinates 6 . 7 Math 130 Linear Algebra 4 5 bnj D Joyce, Fall 2015 Collect these in the columns of a matrix P β to The coordinates of a vector v in a vector space form a transition matrix. V with respect to a basis β = fb1; b2;:::; vbg are 2 3 b11 b12 : : : b1n those coefficients ci which uniquely express v as as 6b b : : : b 7 linear combination of the basis vectors 6 21 22 2n7 P β = 6 . .. 7 4 . 5 v = v1b1 + v2b2 + ··· + vnbn: bn1 bn2 : : : bnn These coefficients v1; v2; : : : ; vn are called coordi- Now, the β-coordinates for bj are all 0s except a th nates with respect to the basis β. The column vec- 1 in the j coordinate, so multiplying P β by the th tor of these coordinates is denoted [v]β. column matrix [bj]β picks out the j column, which are the standard coordinates for bj, so P β[bj]β = 2 3 v1 [bj]. 6v27 More generally, for an arbitrary vector v in Fn, [v]β = 6 . 7 6 . 7 the β-coordinates [v]β of v as a linear combination 4 5 of the basis vectors in β, so vn P [v] = [v] : When the basis is the standard basis for F n β β Thus, the transition matrix P β converts from = fe1; e2;:::; eng; β coordinates to coordinates. Unfortunately, it's usually the reverse change of then the coordinates [v] of a vector v = coordinates that we want. But we can do that, too. (v1; v2; : : : ; vn) are just the usual coordinates of v. To convert the other way, just invert the matrix P to get 2v 3 β 1 P = (P )−1: 6v 7 β β 6 27 [v] = 6 . 7 Then, 4 . 5 Pβ [v] = [v]β: vn Example 1. A low-dimensional example will help 2 Changing between standard coordinates explain things. Let β = fb1; b2g be a basis of R with respect to another. Take the case when where b1 = (3; 1) and b2 = (−4; 2). V is F n and the basis β is not the standard basis In the figure, the standard coordinates are shown . We may have the standard coordinates of a vec- with black axes and a yellow grid, while the β- tor and want the β coordinates of it, or vice versa. coordinates are shown with blue axes and a cyan How do we convert back and forth? grid. 1 H H H H H H Suppose we had the β-coordinates for a vector, say HH HH H H the vector v = 2b1 + 3b2 + 4b3. Its β-coordinates H H 2 3 H H 2 HH HH H H H are [v] = 3 . We can use P to convert those H H H β 4 5 β H H HH 4 H HH v H HHY H H H H H to standard -coordinates: HH H HH H b2 H 1H H H [v] = P β[v]β H H H b1 H H HH 2−1 1 1 3 223 253 H H H HH HH H = 4 1 −1 1 5 435 = 435 H H H H 1 1 −1 4 1 H H H HH HH H H H Therefore, as a 3-tuple, v is (5; 3; 1). H HH H H The inverse of the matrix P β is the matrix H H H HH P , and it can be computed by the methods de- HH H β H H H H scribed before. You'll find H H H HH 2 3 HH H 0 1=2 1=2 H HH Pβ = 41=2 0 1=25 The transition matrix is 1=2 1=2 0 3 −4 Using that matrix we can convert from standard P = β 1 2 -coordinates to β-coordinates. For example, if we take standard coordinates for the vector v we had and its inverse is 253 before, [v] = 3 , and multiply on the left by 0:2 0:4 4 5 P = (P )−1 = 1 β β −0:1 0:3 Pβ , we should get the β-coordinates we started Take a typical vector v, say v = (2; 3). Then its with β-coordinates are [v]β = Pβ [v] 0:2 0:4 2 1:6 2 0 1=2 1=23 253 223 [v] = P [v] = = β β −0:1 0:3 3 0:7 = 41=2 0 1=25 435 = 435 1=2 1=2 0 1 4 The same vector v in the plane can be described in two ways. In standard coordinates go 2 e1's (2 units Changing between coordinates with respect right) and 3 e2's (3 units up), or in β-coordinates, to two different bases. How do you convert be- go 1.6 b1's and 0.7 b2's. tween coordinates [v]γ of a vector v with respect to Example 2. A three-dimensional case. Let V = a basis γ and coordinates [v]β with respect to a 3 R , and consider the basis β = fb1; b2; b3g where different basis β? b1 = (−1; 1; 1), b2 = (1; −1; 1), and b3 = One way is to do the same as we just did where (1; 1; −1). The transition matrix which converts β- the basis γ replaces the standard basis . Start coordinates to standard -coordinates puts the bj's by finding the coordinates of the basis vectors of in columns β with respect to the basis γ. Then put them in columns in a matrix, which we denote P . And 2 −1 1 1 3 γ β we get analogous results: P β = 4 1 −1 1 5 1 1 −1 [v]γ = Pγ β [v]β 2 but for two different bases γ and β. If you want the Similar matrices and equivalence relations. reverse change of coordinates, invert the matrix. Knowing when two matrices represent the same lin- P = (P )−1: ear operator is so important that there's a name for β γ γ β them. Then, P [v] = [v] : Definition 4. Two square matrices A and B are β γ γ β said to be similar or conjugate when there is an n In the case that the vector space V is F , we can invertible matrix Q such that B = Q−1AQ: We'll use the standard basis as an intermediate step. We denote similar matrices A ∼ B. can also compute Pγ β as a composition With that definition, we can summarize the pre- −1 Pγ β = Pγ P β = (P γ) P β vious theorem as saying similar matrices represent which is easy to use since the columns of P γ and the same linear operator. P β are the basis vectors of γ and of β, respec- Similarity is a binary relation that has three im- tively. portant properties • Reflexivity. Any square matrix is similar to Matrix representation of linear operators. itself. A ∼ A. A linear operator is just a linear transformation T : V ! V from a vector space to itself. In order • Symmetry. If one matrix is similar to another, to represent a linear transformation between two then the other is similar to it. A ∼ B implies different vector spaces, you need to choose a basis B ∼ A. for each, but for linear operators, only one basis β • Transitivity. If one matrix is similar to an- for V is needed. If you choose a different basis γ other, and the second is similar to the third, for V , you'll get a different matrix. β then the first is similar to the third. A ∼ B If the matrix [T ]β represents T for the basis β, γ and B ∼ C imply A ∼ C. how can you find the matrix [T ]γ that represents T for the basis γ? Answer, just use the two change of Any binary relation that has these three proper- ties is called an equivalence relation. coordinate matrices Pγ β and Pγ β. Equivalence relations occur throughout mathe- If you start with a vector [v]γ in γ-coordinates, matics. You're familiar with a few of them. For ex- first hit it with Pβ γ to get it in β-coordinates. β ample, similarity of triangles in geometry. Also con- Now you've got [v]β, so hit that with [T ] to get β gruence of triangles. In calculus, having the same [T (v)]β. Finally, hit that with Pγ β to get [T (v)]γ. Thus, derivative is an equivalence relation although it's γ β usually not called an equivalence relation in a cal- [T ]γ = Pγ β[T ]βPβ γ: −1 culus course. In number theory, congruence modulo Since (Pγ β) = Pβ γ, we can also write that equa- n is an equivalence relation. tion as γ −1 β Outside of mathematics equivalence relations are [T ]γ = Pβ γ[T ]βPβ γ: common, too. Being the same height, being on the This observation yields the following theorem same basketball team in a sports league, and hav- where the matrix Q in the statement is the transi- ing the same parents are three different equivalence tion matrix Pβ γ[T ]. relations. Theorem 3. Two matrices A and B represent the In fact, the word \same" indicates there's an as- same linear operator if and only if there is an in- sociated equivalence relation.