1. Determinants
CHANGE OF BASIS
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2. 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= Find 𝑥 𝐶.
(1)
(2)
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3. 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)
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4. CHANGE OF BASIS
This formula gives 𝑥 𝐶, once we know the columns of the matrix. From (1),
b1 𝐶= and b2 𝐶= −6 1
Thus, (3) provides the solution:
𝑥 𝐶= 4 − = 6 4
The C-coordinates of x match those of the x in Fig. 1, as seen on the next slide.
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5. CHANGE OF BASIS
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6. 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)
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7. 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).
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8. 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)
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9. 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 b 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.
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10. 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
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11. 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 = −4 −5 ⋮ −9 −5 1 −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)
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