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91Ó°ÊÓ

a.Find the change of basis matrix S from the basis B considered in Exercise 28 to the standard basisU=(1,t,t2)ofP2considered in Exercise 27.

b.Verify the formula SB=AS for the matrices B and A you found in Exercises 28 and 27,respectively.

c.Find the change of basis matrix from U to B

Short Answer

Expert verified

a.The solution isS=(1−1101−2001)

b. Verified

c.The solution isS−1=(111012001)

Step by step solution

01

Step 1:Definition for the B matrix of transformation T

Consider a linear transformation T from V to V, where V is an n-dimensional linear space .Let B be a basis of V.

Consider the linear transformationLB∘T∘LB−1 fromRntoRn with standard matrix B

Which impliesBx→=LB[T(LB−1(x→))]∶Äx→ â¶Ä‰in â¶Ä‰Rn

This matrix B is called the B matrix of transformation T.

02

Step 2:Definition for the change of basis matrix of transformation T

Consider two bases U and B of an n-dimensional linear space V.

Consider the linear transformationLU∘LB−1 from Rnto Rnwith standard matrix S,meaning thatSx→=LU[LB−1(x→)] â¶Ä‰âˆ¶Ä â¶Ä‰â€‰x→ â¶Ä‰in â¶Ä‰Rn

This invertible matrix S is called the change of basis matrix from B to U,sometimes denoted bySB→U

03

(a) Solution for the change of matrix S

Consider the bases as B and U as follows

B=[1 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰âˆ’1 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰10 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰2 â¶Ä‰â€‰â¶Ä‰â€‰âˆ’40 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰0 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰4]U=[1 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰âˆ’1 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰10 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰2 â¶Ä‰â€‰â¶Ä‰â€‰âˆ’40 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰0 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰4]

Now by inspection we find out the change of matrix from B to U

SB→U=[a−b+cb−2cc]SB→U=(1−1101−2001)

Hence the solution.

04

(b) Solution for the SB=AS

Consider the matrix as follows

B=[1 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰âˆ’1 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰10 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰2 â¶Ä‰â€‰â¶Ä‰â€‰âˆ’40 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰0 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰4]A=[1 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰âˆ’1 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰10 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰2 â¶Ä‰â€‰â¶Ä‰â€‰âˆ’40 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰0 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰4]

Here both the A and B matrix is same.

S=(1−1101−2001)

SB=(1−1102−4004)[1 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰âˆ’1 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰10 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰1 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰âˆ’20 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰0 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰1]SB=(1−2502−8004)

Similarly,AS becomes as follows

AS=[1 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰âˆ’1 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰10 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰1 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰âˆ’20 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰0 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰1](1−1102−4004)AS=(1−2502−8004)

Hence the proof

05

(c) Solution for the change of matrix from U to B

Consider the bases as B and U as follows

B=[1 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰âˆ’1 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰10 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰2 â¶Ä‰â€‰â¶Ä‰â€‰âˆ’40 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰0 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰4]U=[1 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰âˆ’1 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰10 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰2 â¶Ä‰â€‰â¶Ä‰â€‰âˆ’40 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰0 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰4]

Now by inspection we find out the change of matrix from U to B

S−1U→B=[a+b+cb+2cc]S−1U→B=(111012001)

Hence the solution.

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