Determinants CHANGE OF BASIS © 2016 Pearson Education, Inc.

CHANGE OF BASIS Example 1 Consider two bases 𝛽= {b1, b2} and C = {c1, c2} for a vector space V, such that 𝑏1=4𝑐1+𝑐2 and 𝑏2=−6𝑐1+𝑐2 Suppose 𝑥=3𝑏1+𝑏2 That is, suppose 𝑥 𝛽= 3 1 . Find 𝑥 𝐶. (1) (2) © 2016 Pearson Education, Inc.

CHANGE OF BASIS Solution Apply the coordinate mapping determined by C to x in (2). Since the coordinate mapping is a linear transformation, 𝑥 𝐶= 3b1+b2 𝐶 = 3b1]𝐶 +[b2 𝐶 We can write the vector equation as a matrix equation, using the vectors in the linear combination as the columns of a matrix: 𝑥 𝐶= b1 𝐶 b2 𝐶 3 1 (3) © 2016 Pearson Education, Inc.

CHANGE OF BASIS This formula gives 𝑥 𝐶, once we know the columns of the matrix. From (1), b1 𝐶= 3 1 and b2 𝐶= −6 1 Thus, (3) provides the solution: 𝑥 𝐶= 4 −6 1 1 3 1 = 6 4 The C-coordinates of x match those of the x in Fig. 1, as seen on the next slide. © 2016 Pearson Education, Inc.

CHANGE OF BASIS © 2016 Pearson Education, Inc.

CHANGE OF BASIS Theorem 15: Let 𝛽= {b1, . . . , bn} and C = {c1, . . . , c2} for a vector space V . Then there is a unique n × n matrix c 𝑃 𝛽 such that 𝑥 𝐶= c 𝑃 𝛽 𝑥 𝛽 The columns of c 𝑃 𝛽 are the C-coordinate vectors of the vectors in the basis 𝛽. That is, c 𝑃 𝛽 =[ b1 𝐶 b2 𝐶 . . . bn 𝐶] (4) (5) © 2016 Pearson Education, Inc.

CHANGE OF BASIS The matrix c 𝑃 𝛽 in Theorem 15 is called the change-of-coordinates matrix from 𝛽 to C. Multiplication by c 𝑃 𝛽 converts 𝛽-coordinates into C-coordinates. Figure 2 below illustrates the change-of-coordinates equation (4). © 2016 Pearson Education, Inc.

CHANGE OF BASIS The columns of c 𝑃 𝛽 are linearly independent because they are the coordinate vectors of the linearly independent set 𝛽. Since c 𝑃 𝛽 is square, it must be invertible, by the Invertible Matrix Theorem. Left-multiplying both sides of equation (4) by (c 𝑃 𝛽)-1 yields (c 𝑃 𝛽)-1 𝑥 𝐶 = 𝑥 𝛽 Thus (c 𝑃 𝛽)-1 is the matrix that converts C-coordinates into 𝛽–coordinates. That is, (c 𝑃 𝛽)-1 = 𝛽 𝑃 C (6) © 2016 Pearson Education, Inc.

CHANGE OF BASIS IN ℝ 𝑛 If 𝛽= {b1, . . . , bn} and ℰ is the standard basis {e1, . . . , en} in ℝ 𝑛 , then b1 ℰ= b1,and likewise for the other vectors in 𝛽. In this case, ℰ 𝑃 𝛽 is thesame as the change-of-coordinates matrix P𝛽 introduced in Section 4.4, namely, P𝛽= [ b1 b2 . . . bn] To change coordinates between two nonstandard bases in ℝ 𝑛 ,we need Theorem 15. The theorem shows that to solve the change-of-basis problem, we need the coordinate vectors of the old basis relative to the new basis. © 2016 Pearson Education, Inc.

CHANGE OF BASIS IN ℝ 𝑛 Example 2 Let b1= −9 1 , b2= −5 −1 ,c1= 1 −4 , c2= 3 −5 and consider the bases for ℝ 𝑛 given by 𝛽= {b1, b2} and C = {c1, c2}. Find the change-of-coordinates matrix from 𝛽 to C. Solution The matrix 𝛽 𝑃 C involves the C-coordinate vectors of b1 and b2. Let b1 𝐶= 𝑥1 𝑥2 and b2 𝐶= 𝑦1 𝑦2 .Then, by definition, [𝑐1 𝑐2] 𝑥1 𝑥2 = b1 and [𝑐1 𝑐2] 𝑦1 𝑦2 = b2 © 2016 Pearson Education, Inc.

CHANGE OF BASIS IN ℝ 𝑛 To solve both systems simultaneously, augment the coefficient matrix with b1 and b2, and row reduce: 𝑐1 𝑐2⋮𝑏1 𝑏2 = 1 3 −4 −5 ⋮ −9 −5 1 −1 ~ 1 0 0 1 ⋮ 6 4 −5 −3 Thus b1 𝐶= 6 −5 and b2 𝐶= 4 −3 The desired change-of-coordinates matrix is therefore c 𝑃 𝛽 = 𝑏1 𝑐 𝑏2 𝑐 = 6 4 −5 −3 (7) © 2016 Pearson Education, Inc.