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1
Quiz 7 (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 .
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Question 2: Consider the following bases of R3 and R2 :
S = {w1 , w2 , w3 } = {(1, 1, 1), (1, 1, 0), (1, 0, 0)},
S 0 = {u1 , u2 } = {(1, 3), (2, 5)}.
Let F : R3 −→ R2 be the linear mapping defined by F (x, y, z) = (3x+2y −4z, x−5y +3z). Find
the matrix of F in the bases S and S 0 . (Hint use the fact that (a, b) = (−5a+2b)u1 +(3a−b)u2 .)
We compute the coordinate vectors of F (w1 ), F (w2 ), and F (w3 ) relative to S 0 .
F (w1 ) = (1, −1) = (−5 − 2)u1 + (3 + 1)u2 = −7u1 + 4u2 ,
F (w2 ) = (5, −4) = (−25 − 8)u1 + (15 + 4)u2 = −33u1 + 19u2 ,
F (w3 ) = (3, 1) = (−15 + 2)u1 + (9 − 1)u2 = −13u1 + 8u2 .
This means that
[F ]SS 0 =
−7
4
−33 −13
.
19
8
2
Quiz 7, October 30, 2014
Question 3: Let (V, h·, ·i) be a finite-dimensional inner product space. Let F : V −→ V be a
linear mapping. Assume that F has the following property: hF (v), vi ≥ 5kvk2 for all vectors v
in V . Prove that F is bijective.
Let us characterize Ker(F ). Consider v ∈ Ker(F ). Then F (v) = 0. This implies that 0 =
hF (v), vi ≥ 5kvk2 . This proves that kvk = 0, which in turn proves that v = 0. As a result
Ker(F ) = {0}. This proves that F is injective. Owing to the rank theorem, this also proves that F
is surjective since the dimension of the co-domain is equal to the dimension of the domain. Hence
F is bijective
Question 4: It is possible to find two nonzero square matrices A and B such that tr((A +
1
1
1
B)T (A + B)) 2 > tr(AT A) 2 + tr(B T B) 2 ? (be very specific in your answer)
The mapping (A, B) 7−→ tr(AT B) is an inner product. We have proved it in class. This inner
1
product induces a norm: kAk = tr(AT A) 2 . As a result the triangle inequality holds,
kA + Bk ≤ kAk + kBk.
for all square matrices A and B. Hence it is impossible to find two nonzero square matrices A and
1
1
1
B such that tr((A + B)T (A + B)) 2 > tr(AT A) 2 + tr(B T B) 2 .