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1
Mid term exam 2 (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
Rb
no justification will not be graded. Some useful facts: (f, g) 7−→ a f (t)g(t)dt is an inner
product in C 0 ([a, b]; R) and (A, B) 7−→ tr(AT B) is an inner product for square matrices.
Question 1: Are the following polynomials p1 (t) = 1 − 3t + 2t2 , p2 (t) = 2 − 4t − t2 , p3 (t) =
1 − 5t + 7t2 linearly independent in P2 (t)?
Let X = (x1 , x2 , x3 ) ∈ R3 be so that x1 p1 + x2 p2 + x3 p3 = 0. Then
x1 p1 (t) + x2 p2 (t) + x3 p3 (t) = x1 + 2x2 + x3 + (−3x1 − 4x2 − 5x3 )t + (2x1 − x2 + 7x3 )t2 = 0.
This means that X solves

1
−3
2
2
−4
−1

1
−5 X = 0.
7
Let us reduce the matrix of the linear system in echelon form

 


1
2
1
1 2
1
1
−3 −4 −5 ∼ 0 2 −2 ∼= 0
2 −1 7
0 −5 5
0
2
2
0

1
−2 .
0
There are only two pivots. There is one free variable. This means that the solution set of the above
linear system is not {0}. There is some nonzero vector X so that x1 p1 + x2 p2 + x3 p3 = 0. This
means that p1 , p2 , p3 are linearly dependent.
Question 2: 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 ≥ 2kvk2 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 ≥ 2kvk2 . 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
2
Mid term exam 2, October 26, 2010
1
1
1
Question 3: Is it true that tr((A + B)T (A + B)) 2 ≤ tr(AT A) 2 + tr(B T B) 2 for all square
matrices A and B? If yes, why (be very specific); if no, give a counter-example.
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 . The proposed formula is just the triangle inequality,
kA + Bk ≤ kAk + kBk.
which we know from class is true for all norms.
R
R
12 R
12
e
e
e
2
g(t)
dt
Question 4: Is it true that log π f (t)g(t)dt ≤ log π f (t)2 dt
for all conlog π
tinuous functions? If yes, why (be very specific); if no, give a counter-example.
Re
The mapping (f, g) 7−→ log π f (t)g(t)dt is an inner product. We have proved it in class. This inner
R
12
e
2
k. The proposed formula is just the Cauchyproduct induces a norm: kf k =
f
(t)
dt
log π
Schwarz inequality,
hf, gi ≤ kf kkgk,
which we know from class is true for all inner products.
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Question 5: Find two nonzero linear mappings T : R2 −→ R2 and F : R2 −→ R2 so that
F (T (X)) = 0 for all X in R2
Let A and B two matrices in M2,2 (R) and consider the following mappings:
T (X) = BX,
F (X) = AX.
Then F (T (X)) = F (BX) = ABX. The problem consists of finding A 6= 0 and B 6= 0 so that
AB = 0. Let us consider the following matrices
0 1
A=
,
B=A
0 0
Then
AB = 0
In conclusion the mappings F and T defined above are not zero and are so that F (T (X)) = 0 for
all X in R2 .
Question 6: Let F : P1 (t) −→ P1 (t) be the linear mapping such that F (t) = 1 + 4t and
F (1 + 2t) = 2 + 3t. Find F (a + bt) for all pairs (a, b) in R2 .
Let a + bt in P1 (t). We write a + bt as a linear combination of t, and 1 + 2t. (This is always
possible since {t, 1 + 2t} is clearly a basis of P1 (t).) Let us find the two real numbers x and y so
that a + bt = xt + y(1 + 2t). This is equivalent to saying
a = y,
and b = x + 2y.
This implies
x = b − 2a
and y = a.
Then, the following holds owing to the fact that the mapping F is linear
F (a+bt) = F (xt+y(1+2)t) = xF (t)+yF (1+2t) = (b−2a)(1+4t)+a(2+3t) = b−2a+2a+(4b−8a+3a)t.
In conclusion
F (a + bt) = b + (4b − 5a)t.
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Mid term exam 2, October 26, 2010
Question 7: Consider the set of two-variate polynomials of degree at most one P1 (s, t) =
{as + bt + c; (a, b, c) ∈ R3 }. Consider F : P1 (s, t) −→ R3 the linear mapping defined by
F (p) = (p(0, 0), p(1, 1), p(4, 2)). Show that F is bijective (Hint: observe that dim(P1 (s, t)) = 3).
Since P1 (s, t) and R3 have the same dimension, it suffices to prove that F is injective (by the rank
Theorem). Let p = c + bs + at be a member of P1 (s, t) such that F (p) = 0. Then
0 = p(0, 0) = c,
0 = p(1, 1) = a + b + c 0 = p(4, 2) = 4a + 2b + c.
This is a linear system for a, b and c.

4
1
0
The associated
 
2 1
4
1 1 ∼ 0
0 1
0
matrix is
2
−2
0

1
−3
1
The rank is maximum. This means that a = b = c = 0. As a result F is injective. In conclusion F
is bijective.
Question 8: Let V = C 0 ([−π, π]; R) be the vector space over R composed of Rthe functions
π
from [−π, π] to R that are continuous. Consider the inner product hf, gi = π1 −π f (t)g(t)dt
√
and√the associated norm k · k. We recall that k1/ 2k = k sin(t)k = k√cos(t)k = 1 and B =
{1/ 2, sin(t),
√ cos(t)} are orthonormal. Let U
√ = span(B). Let u1 (t) = 1/ 2 + 2 sin(t) + 4 cos(t),
u2 (t) = 2/ 2 − 3 sin(t) + cos(t), u3 (t) = 2/ 2 + sin(t) − cos(t) in V . (a) Show that u1 , u2 , u3
are orthogonal and compute ku1 k, ku2 k, ku3 k.
We compute hui , uj i for i 6= j, i, j ∈ {1, 2, 3}.
√
hu1 , u2 i = 2k1/ 2k − 6k sin k + 4k cos k = 2 − 6 + 4 = 0,
√
hu1 , u3 i = 2k1/ 2k + 2k sin k − 4k cos k = 2 + 2 − 4 = 0,
√
hu2 , u3 i = 4k1/ 2k − 3k sin k − 1k cos k = 4 − 3 − 1 = 0.
This proves the statement. Clearly
ku1 k =
√
21,
ku2 k =
√
14,
ku3 k =
√
6.
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(b) What is the dimension of U = span(B)? Is C = {u1 (t), u2 (t), u3 (t)} a basis of U ? (why).
The vectors composing B are linearly independent since they are orthogonal. Moreover B is a
spanning set of U by definition. Hence B is a basis of U . As a result dim(U ) = 3.
C is a basis of U since (1) it is a subset of U , (2) the vectors composing C are linearly independent
(since they are orthogonal) and (3) the cardinal number of C equals the dimension of U .
√
(c) Consider the linear mapping F : U −→ U so that F (x/ 2 + y sin(t) + z cos(t)) = xu1 (t) +
yu2 (t) + zu3 (t) for all x, y, z ∈ R3 . Consider the base C = {u1 (t), u2 (t), u3 (t)}. Compute the
matrices [F ]B,B and [F ]B,C ?
Observe that
√
√
F (1/ 2) = u1 (t) = 1/ 2 + 2 sin(t) + 4 cos(t),
√
F (sin(t)) = u2 (t) = 2/ 2 − 3 sin(t) + cos(t),
√
F (cos(t)) = u3 (t) = 2/ 2 + sin(t) − cos(t)
This means that
[F ]B,B

1
= 2
4
2
−3
1

2
1
−1
and
[F ]B,C

1
= 0
0
0
1
0

0
0
1
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Mid term exam 2, October 26, 2010
(d) Prove that the linear mapping F defined in question (c) is bijective.
There are many ways to prove this. The easiest answer is that F is bijective because the matrix
[F ]B,C is clearly invertible. Another way to prove the statement √
consists of characterizing Ker(F ).
Let v ∈ Ker(F ). There are numbers x, y, z in R3 so that v = x/ 2 + y sin(t) + z cos(t) since B is
a basis of U . Then
F (v) = 0 = xu1 (t) + yu2 (t) + zu3 (t),
which implies
hF (v), u1 i = xku1 k2 = 0,
hF (v), u2 i = yku2 k2 = 0,
hF (v), u3 i = zku3 k2 = 0.
As a result x = y = z = 0, which in turn implies v = 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.
√
(e) Let v(t) = 3/ 2 + 5 sin(t) + 2 cos(t). Find u(t) ∈ U so that F (u) = v (Hint: use the fact
that u1 , u2 , u3 are orthogonal).
√
Let x1 , x2 , x3 ∈ R3 so that u = x1 / 2 + x2 sin(t) + x3 cos(t). By definition of F
√
v = F (u) = x1 F (1/ 2) + x2 F (sin(t)) + x3 F (cos(t)) = x1 u1 + x2 u2 + x3 u3
Taking the inner product with u1 and using the orthogonality of u1 , u2 , u3 gives
hv, u1 i = 21 = x1 hu1 , u1 i + x2 hu2 , u1 i + x3 hu3 , u1 i = x1 ku1 k2 = 21x1 ,
i.e., x1 = 1. Similarly
hv, u2 i = −7 = x1 hu1 , u2 i + x2 hu2 , u2 i + x3 hu3 , u2 i = x2 ku2 k2 = 14x2 ,
i.e., x1 = − 21 .
hv, u3 i = 9 = x1 hu1 , u3 i + x2 hu2 , u3 i + x3 hu3 , u3 i = x3 ku3 k2 = 6x3 ,
i.e., x1 = 23 . Hence the solution is
In other words
 
 
1
x1
x2  . = − 1  .
2
3
x3
2
√
3
1
v(t) = 1/ 2 − sin(t) + cos(t)
2
2