Analysis & Group theory problems
Solutions
1. Find the minimum value of f (x) = (2x2 − 5x + 2)3 over the closed interval [0, 1].
• Since f is continuous on a closed interval, it suffices to check the endpoints, the points
at which f 0 does not exist and the points at which f 0 is equal to zero. In this case,
f 0 (x) = 3(2x2 − 5x + 2)2 · (2x2 − 5x + 2)0
= 3(2x2 − 5x + 2)2 · (4x − 5)
is zero when x = 5/4 and also when the quadratic factor is zero, namely when
√
5±3
5 ± 25 − 4 · 2 · 2
=
=⇒ x = 2, x = 1/2.
x=
2·2
4
Since x = 5/4 and x = 2 do not lie in the given closed interval, this means that
x = 0,
x = 1,
x = 1/2
are the only points at which the minimum value may occur. Once we now compute
f (1) = −1,
f (0) = 8,
f (1/2) = 0,
we may finally conclude that the minimum value is f (1) = −1.
2. Show that log x ≤ x − 1 for all x > 0.
• Letting f (x) = log x − x + 1 for convenience, one easily finds that
1
1−x
−1=
.
x
x
Thus, f 0 (x) is positive if and only if 1 − x > 0, hence if and only if x < 1. This shows
that f is increasing when x < 1 and also decreasing when x > 1, so
f 0 (x) =
max f (x) = f (1) = log 1 − 1 + 1 = 0
=⇒
f (x) ≤ max f (x) = 0.
3. Suppose that x > y > 0. Using the mean value theorem or otherwise, show that
x
y
1 − < log x − log y < − 1.
x
y
• Letting f (x) = log x for convenience, we use the mean value theorem to find that
f 0 (c) =
f (x) − f (y)
x−y
=⇒
1
log x − log y
=
c
x−y
for some y < c < x. Inverting these positive numbers reverses the inequality, hence
1
1
1
< <
x
c
y
=⇒
=⇒
1
log x − log y
1
<
<
x
x−y
y
y
x
1 − < log x − log y < − 1.
x
y
4. Let ∗ be an operation on a set S. Suppose that ∗ is associative, and that S contains
an identity element for ∗. Prove that if x, y are two elements of S which are invertible
with respect to ∗, then x ∗ y is invertible with respect to ∗. What is (x ∗ y)−1 ?
• Let e denote the identity element for ∗. Since x and y are both invertible, they have
inverses x−1 and y −1 in S. Let a = x ∗ y and let b = y −1 ∗ x−1 . Using associativity, we
have
a∗b = (x∗y)∗b = x∗(y∗b) = x∗(y∗(y −1 ∗x−1 )) = x∗((y∗y −1 )∗x−1 ) = x∗(e∗x−1 ) = x∗x−1 = e
and, similarly,
b∗a = b∗(x∗y) = (b∗x)∗y = ((y −1 ∗x−1 )∗x)∗y = (y −1 ∗(x−1 ∗x))∗y = (y −1 ∗e)∗y = y −1 ∗y = e.
So a ∗ b = e = b ∗ a, hence a = x ∗ y is invertible, with inverse a−1 = b. Hence
(x ∗ y)−1 = y −1 ∗ x−1 .
5. Let GL(2, R) denote the set of invertible 2 × 2 matrices with real entries. Which of the
following is a group operation on GL(2, R)?
(a). matrix addition
(b). matrix multiplication
(c). the operation ∗ given by A ∗ B = AB − BA for matrices A and B
• (a) Let I denote the 2 × 2 identity matrix. Then I, −I ∈ GL(2, R) but I + (−I) = 0 6∈
GL(2, R). So GL(2, R) is not closed under addition. So addition is not an operation
on GL(2, R), so it is certainly not a group operation on GL(2, R).
(b) Matrix multiplication is a group operation on GL(2, R). Indeed:
– If A is any 2 × 2 matrix with real entries, then A ∈ GL(2, R) if and only if
det(A) 6= 0. So, if A, B ∈ GL(2, R) then det(A) 6= 0 and det(B) 6= 0. Hence
det(AB) = det(A) det(B) 6= 0, so AB ∈ GL(2, R). This proves that A, B ∈
GL(2, R) =⇒ AB ∈ GL(2, R), so matrix multiplication is an operation on
GL(2, R).
– Matrix multiplication is associative (I guess this is proven in MA1111; otherwise,
it’s a simple calculation).
– The 2 × 2 identity matrix I satisfies I ∈ GL(2, R) and IA = AI = A for
any A ∈ GL(2, R). So matrix multiplication on GL(2, R) has an identity element, namely I.
– If A ∈ GL(2, R) then A is an invertible matrix, so there is a matrix B (often
written B = A−1 ) such that AB = BA = I. So A is invertible with respect to
matrix multiplication.
(c) GL(2, R) is not closed under ∗, since, for example, I ∈ GL(2, R) and I ∗ I =
I 2 − I 2 = 0 6∈ GL(2, R). So ∗ is not an operation on GL(2, R), so it is certainly not a
group operation on GL(2, R).