math 216 Homework 5.1 PDF

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MATH 107.01 HOMEWORK #15 SOLUTIONS

Problem 5.1.3. Determine if T : R3 → R3 given by     x x+y+z T  y  = z − y − x  xyz z is linear.

Solution. Note   1 T  0 + T 1

that                 0 1 1 4 4 2 2 0 1 =  0 +  0 =  0 6= 0 = T  1 = T  0 +  1. 1 1 2 2 0 0 0 1

Hence T is not linear.



Problem 5.1.4. Determine if T : R3 → R2 given by     x 2x − 2y + 5z T y  = x + 2z z is linear.

Solution. This is T (X) = AX where  2 −2 A= 1 0

 5 . 2

Hence T is linear.



Problem 5.1.7. Determine if T : P2 → P2 given by   2 T ax2 + bx + c = a (x + 1) + b (x + 1) + c

is linear.

Solution. Note that      T λ1 a1 x2 + b1 x + c1 + λ2 a2 x2 + b2 + c2   = T (λ1 a1 + λ2 a2 ) x2 + (λ1 b1 + λ2 b2 ) x + (λ1 c1 + λ2 c2 ) 2

= (λ1 a1 + λ2 a2 ) (x + 1) + (λ1 b1 + λ2 b2 ) (x + 1) + (λ1 c1 + λ2 c2 ) 2

= λ1 a1 (x + 1) + λ1 b1 (x + 1) + λ1 c1 2

+ λ2 a2 (x + 1) + λ2 b2 (x + 1) + λ2 c2  2 = λ1 a1 (x + 1) + b1 (x + 1) + c1   2 + λ2 a2 (x + 1) + b2 (x + 1) + c2     = λ1 T a1 x2 + b1 x + c1 + λ2 T a2 x2 + b2 x + c2 

1

2

MATH 107.01 HOMEWORK #15 SOLUTIONS

whenever λ1 , λ2 ∈ R and a1 x2 + b1 x + c1 , a2 x2 + b2 + c2 ∈ P2 . Hence T is linear.  Problem 5.1.12. Determine if T : Mn×n (R) → R given by T (A) = det (A) is linear. Proof. Note that T (I + I) = T (2I ) = det (2I ) = 2n det (I ) = 2n 6= 2 = det (I )+det (I ) = T (I)+T (I) whenever n ≥ 2. Hence T is not linear.



Problem 5.1.13. Recall that the collection R+ of positive real numbers is a vector space under the addition x ⊕ y = xy and the scalar multiplication λ ⊙ x = xλ . (a) Show that the natural logarithm is a linear transformation ln : R+ → R. (b) Show that the exponential map is a linear transformation e : R → R+ . Solution. (a) Note that         ln (λ1 ⊙ x1 ⊕ λ2 ⊙ x2 ) = ln x1λ1 ⊕ xλ22 = ln x1λ1 x2λ2 = ln xλ11 + ln xλ2 2 = λ1 ln (x1 ) + λ2 ln (x2 )

whenever λ1 , λ2 ∈ R and x1 , x2 ∈ R+ . Hence ln is linear. (b) Note that λ1

eλ1 x1 +λ2 x2 = eλ1 x1 eλ2 x2 = eλ1 x1 ⊕ eλ2 x2 = (ex1 )

λ2

⊕ (ex2 )

= λ1 ⊙ ex1 ⊕ λ2 ⊙ ex2

whenever λ1 , λ2 ∈ R and x1 , x2 ∈ R. Hence e is linear.



Problem 5.1.18. Find a matrix A that expresses the linear transformation T : R4 → R3 given by     x1 x 1 − x 2 + 3x 3 − x 4  x2     T  x3 = 2x1 + 3x2 − x3 − 2x4 3x1 + 7x2 − 5x3 − 3x4 x4 in the form of a matrix transformation T (X) = AX . Solution. Here, A ∈ M3×4 (R) satisfies   1 Ae1 = T e1 =  2 3   3 Ae3 = T e3 =  −1 −5

  −1 Ae2 = T e2 =  3  7   −1 Ae4 = T e4 = −2 . −3

This gives



1 A = 2 3

−1 3 7

3 −1 −5

 −1 −2 . −3

Problem 5.1.20. Let T : R3 → R4 be a linear transformation such that             1 2 1 0 1 1 0   1  0           T −1 =   −1  , T 0 =  0 , T 1 =  0  . −1 1 0 −1 0 0



MATH 107.01

HOMEWORK #15 SOLUTIONS

3

  x (b) Find T y . z Solution. (b) First, note that  1  −1 , 0 

 0  1  −1

  1  0 , 1

form a basis for R3 . Furthermore, since    1 1 0 1 1 =  0 rref −1 0 0 1 −1 0

0 1 0



0 0 1

 x/2 − y/2 − z/2 x/2 + y/2 + z/2 , x/2 + y/2 − z/2

we have         0 1 1 x 1 1 1  y = (x − y − z) −1  + (x + y + z) 0  + (x + y − z)  1  . (1) 2 2 2 −1 1 0 z

Applying T to (1) then gives         0 1 1 x 1 1 1 T  y  = (x − y − z) T  −1 + (x + y + z) T  0 + (x + y − z) T  1  = 2 2 2 1 −1 z 0

so that

        1 2 1 x 0  1 1 0  1 1        T y = (x − y − z)   + (x + y + z)   + (x + y − z)   0  0 −1 2 2 2 z −1 0 0   2x + y  x/2 + y/2 + x/2   =  −x/2 + y/2 + z/2 . −x/2 − y/2 + z/2

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