# Last name: name: 1 Quiz 5 (Notes, books, and calculators are not authorized)

```Last name:
name:
1
Quiz 5 (Notes, books, and calculators are not authorized)
Show all your work in the blank space you are given on the exam sheet. Always justify your
answer. Answers with no justification will not be graded. P2 is the vector space over R of
polynomials of degree at most two.
Question 1: Consider the following two bases of P2 : E = {(1 − t), t, 4t(1 − t)} and F =
{(1 − 2t)(1 − t), t(2t − 1), 4t(1 − t)}. Compute the change of basis matrix PF →E .
Let e1 = 1 − t, e2 = t, e3 = 4t(1 − t), f1 = (1 − 2t)(1 − t), f2 = t(2t − 1), f3 = 4t(1 − t). The
definition of the matrix PF →E is
PF →E = [[f1 ]E [f2 ]E [f3 ]E ] .
We have
1
f1 = (1 − 2t)(1 − t) = (1 − t) − 2t(1 − t) = e1 − e3
2
1
f2 = t(2t − 1) = t + 2t(t − 1) = e2 − e3
2
f3 = e3 .
As a result

PF →E
1
= 0
− 21
0
1
− 12

0
0 .
1
Question 2: Consider the following basis of P2 : C = {1, t, t2 }. Let F : P2 −→ P2 be the linear
Rt
mapping such that F (p) = p(t) + 61 t2 p0 (t) − 0 p(τ )dτ for all p ∈ P2 . Compute [F ]C,C .
Let us denote c1 = 1, c2 = t, c3 = t2 . The definition of [F ]C,C is
[F ]C,C = [[F (c1 )]C [F (c2 )]C [F (c3 )]C ] .
We have
Z t
Z t
1 2 0
F (c1 ) = c1 (t) + t c1 (t) −
c1 (τ )dτ = 1 −
dτ = 1 − t.
6
0
0
Z t
Z t
1 2
1
1 2 0
c2 (τ )dτ = t + t −
τ dτ = t − t2 .
F (c2 ) = c2 (t) + t c2 (t) −
6
6
3
0
0
Z t
Z t
1 2 0
1
F (c1 ) = c3 (t) + t c3 (t) −
c3 (τ )dτ = t2 + t3 −
τ 2 dτ = t2 .
6
3
0
0
As a result

[F ]C,C
1
= −1
0
0
1
− 31

0
0 .
1
2
Quiz 5, October 19, 2010
2
Question 3: Consider the following
basis

 of P2 : C = {1, t, t }. Let G : P2 −→ P2 be the linear
1 0 0
mapping such that [G]C,C =  1 1 0. Let p = a + bt + ct2 be an arbitrary vector in P2 .
1
1
1
3
3
Compute G(p).
The definition of [G]C,C is such that
[G(p)]C = [G]C,C [p]C .
As a result

1
[G(p)]C =  1
1
3
This means that
0
1
1
3
  

0 a
a
0  b  =  a + b  .
1
1
c
1
3a + 3b + c
1
1
G(p) = a + (a + b)t + ( a + b + c)t2 .
3
3
Question 4: Let H : P2 −→ P2 be the linear mapping such that H(a + bt + ct2 ) = a + (a +
b)t + ( 31 a + 13 b + c)t2 . Let F be the mapping defined in Question 2. Compute F (H(p)), for all
p = a + bt + ct2 .
(Long answer): The definitions of F and H imply that
Z t
1
F (H(p)) = H(p) + t2 (H(p))0 −
H(τ )dτ
6
0
1
1
1
1
1
=a + (a + b)t + ( a + b + c)t2 + t2 (a + b + 2( a + b + c)t)
3
3
6
3
3
1
1 1
1
− at − (a + b)t2 − ( a + b + c)t3
2
3 3
3
= a + bt + ct2 = p
In conclusion F (H(p)) = p, i.e., H is the inverse of F .
(Short answer): Observe that G and H are the same operators. Observe also that [F ]C [H]C =
[F ]C [G]C = I, as a result F (G(p)) = p.
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