4 Determinants
4.7
CHANGE OF BASIS
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CHANGE OF BASIS
Example 1 Consider two bases 𝛽𝛽= {b1, b2} and C =
{c1, c2} for a vector space V, such that
𝑏𝑏1 = 4𝑐𝑐1 + 𝑐𝑐2 and 𝑏𝑏2 = −6𝑐𝑐1 + 𝑐𝑐2 (1)
Suppose
𝑥𝑥 = 3𝑏𝑏1 + 𝑏𝑏2 (2)
3
That is, suppose 𝑥𝑥 𝛽𝛽 = . Find 𝑥𝑥 𝐶𝐶 .
1
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CHANGE OF BASIS
Solution Apply the coordinate mapping determined by
C to x in (2). Since the coordinate mapping is a linear
transformation,
𝑥𝑥 𝐶𝐶 = 3b1 + b2 𝐶𝐶
=3 3b1]𝐶𝐶 + [b2 𝐶𝐶
We can write the vector equation as a matrix equation,
using the vectors in the linear combination as the
columns of a matrix:
3
𝑥𝑥 𝐶𝐶 = b1 𝐶𝐶 b2 𝐶𝐶 (3)
1
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CHANGE OF BASIS
This formula gives 𝑥𝑥 𝐶𝐶 , once we know the columns of
the matrix. From (1),
3 −6
b1 𝐶𝐶 = and b2 𝐶𝐶 =
1 1
Thus, (3) provides the solution:
4 −6 3 6
𝑥𝑥 𝐶𝐶 = =
1 1 1 4
The C-coordinates of x match those of the x in Fig. 1,
as seen on the next slide.
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CHANGE OF BASIS
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CHANGE OF BASIS
Theorem 15: Let 𝛽𝛽= {b1, . . . , bn} and C = {c1, . . . , cn2}
for a vector space V . Then there is a unique n × n
𝑃𝑃
matrix c←𝛽𝛽 such that
𝑃𝑃
𝑥𝑥 𝐶𝐶 = c←𝛽𝛽 𝑥𝑥 𝛽𝛽 (4)
𝑃𝑃
The columns of c←𝛽𝛽 are the C-coordinate vectors of
the vectors in the basis 𝛽𝛽. That is,
𝑃𝑃
c←𝛽𝛽 =[ b1 𝐶𝐶 b2 𝐶𝐶 ... bn 𝐶𝐶 ] (5)
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CHANGE OF BASIS
𝑃𝑃
The matrix c←𝛽𝛽 in Theorem 15 is called the change-
of-coordinates matrix from 𝛽𝛽 to C. Multiplication by
𝑃𝑃
c←𝛽𝛽 converts 𝛽𝛽-coordinates into C-coordinates.
Figure 2 below illustrates the change-of-coordinates
equation (4).
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CHANGE OF BASIS
𝑃𝑃
The columns of c←𝛽𝛽 are linearly independent because
they are the coordinate vectors of the linearly
independent set 𝛽𝛽.
𝑃𝑃
Since c←𝛽𝛽 is square, it must be invertible, by the
Invertible Matrix Theorem. Left-multiplying both
𝑃𝑃
sides of equation (4) by (c←𝛽𝛽)-1 yields
𝑃𝑃
(c←𝛽𝛽)-1 𝑥𝑥 𝐶𝐶 = 𝑥𝑥 𝛽𝛽
𝑃𝑃
Thus (c←𝛽𝛽)-1 is the matrix that converts C-coordinates
into 𝛽𝛽–coordinates. That is,
𝑃𝑃 𝑃𝑃
(c←𝛽𝛽)-1 =𝛽𝛽 ←C (6)
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CHANGE OF BASIS IN ℝ𝑛𝑛
If 𝛽𝛽= {b1, . . . , bn} and ℰ is the standard basis {e1, . . . , en}
in ℝ𝑛𝑛 , then b1 ℰ = b1, and likewise for the other vectors
𝑃𝑃
in 𝛽𝛽. In this case, ℰ ← 𝛽𝛽 is thesame as the change-of-
coordinates matrix P𝛽𝛽 introduced in Section 4.4, namely,
P𝛽𝛽= [ b1 b2 . . . bn]
To change coordinates between two nonstandard bases in
ℝ𝑛𝑛 ,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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CHANGE OF BASIS IN ℝ𝑛𝑛
−9 −5 1
Example 2 Let b1 = , b2 = , c1 = , c2 =
1 −1 −4
3
and consider the bases for ℝ𝑛𝑛 given by 𝛽𝛽= {b1, b2}
−5
and C = {c1, c2}. Find the change-of-coordinates
matrix from 𝛽𝛽 to C.
𝑃𝑃
Solution The matrix 𝛽𝛽 ←C involves the C-coordinate
𝑥𝑥1
vectors of b1 and b2. Let b1 𝐶𝐶 = and b2 𝐶𝐶 =
𝑥𝑥2
𝑦𝑦1
𝑦𝑦2 .Then, by definition,
𝑥𝑥1 𝑦𝑦1
[𝑐𝑐1 𝑐𝑐2] = b1 and [𝑐𝑐1 𝑐𝑐2] 𝑦𝑦 = b2
𝑥𝑥2 2
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CHANGE OF BASIS IN ℝ𝑛𝑛
To solve both systems simultaneously, augment the
coefficient matrix with b1 and b2, and row reduce:
1 3 −9 −5 1 0 6 4
𝑐𝑐1 𝑐𝑐2 ⋮ 𝑏𝑏1 𝑏𝑏2 = ⋮ ~ ⋮ (7)
−4 − 5 1 −1 0 1 −5 −3
Thus
6 4
b1 𝐶𝐶 = and b2 𝐶𝐶 =
−5 −3
The desired change-of-coordinates matrix is therefore
𝑃𝑃 6 4
c←𝛽𝛽 = 𝑏𝑏1 𝑐𝑐 𝑏𝑏2 𝑐𝑐 =
−5 −3
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