Solution of the First HKUST Undergraduate Math Competition – Senior Level

1. Note y = f(ex ) ⇔ yy = ex ⇔ y ln y = x. Then dx = (ln y + 1)dy. So

Z Z Z  y2
x=e
x
y=e y=e
y2  y=e y2 y=e 3e2 − 1
f(e ) dx = y(ln y + 1) dy = (y ln y + y) dy = ln y − + = .
x=0 y=1 y=1 2 4 y=1 2 y=1 4

2. (From linear algebra, the inequalities rank(XY ) ≤ rank(X) and rank(XY Z) ≤ rank(Y ) are useful.)
(Solution 1) Since the first two rows of AB are linearly independent, so 2 ≤ rank(AB) ≤ rank(A) ≤ 2.
Hence rank(AB) = 2.
Next to get BA, we note rank(BA) ≥ rank(A(BA)B) = rank((AB)2 ). Now
 2  
8 2 −2 72 18 −18
(AB)2 =  2 5 4  =  18 45 36  = 9AB.
−2 4 5 −18 36 45

Since BA is a 2 × 2 matrix and rank(9AB) = 2, so rank(BA) = 2. Hence BA is invertible. Finally

(BA)3 = B(ABAB)A = B(AB)2 A = 9BABA = 9(BA)2 . Cancelling (BA)2 , we get BA = 9I.

(Solution 2 due to Lau Lap Ming) Since the first two rows of AB are linearly independent, so 2 ≤
rank(AB) ≤ rank(A) ≤ 2. Then rank(AB) = rank(A) = 2.
Next det(AB − tI) = −t(t − 9)2 , so the eigenvalues of AB are 0 and 9. If λ is an eigenvalue of BA
with eigenvector v 6= 0, then AB(Av) = A(BAv) = A(λv) = λAv. Since A is 3 × 2 and of rank 2, A is
injective. Hence, Av 6= 0 and λ is an eigenvalue of AB. This implies the only possible eigenvalues of BA
are 0 or 9. From
 row operations on the matrix of AB, we see the eigenspace of AB for theeigenvalue
 0is
1 2 −2
spanned by  −2  and the eigenspace V of AB for the eigenvalue 9 is spanned by  1  and  0  .
2 0 1
Restricting to V, AB : V → R2 → V is bijective since AB = 9I on V. So the linear maps B : V → R2
and A : R2 → V must be bijective. In particular, B(V ) = R2 . Then for every x ∈ R2 , there exists v ∈ V
such that Bv = x. So we have BAx = B(ABv) = B(9v) = 9Bv = 9x. Therefore, BA = 9I.

3. (This is an existence problem with solution to be found among continuous functions on [0, 1]. In a
course on metric spaces, a key theorem on existence problem is the contractive mapping theorem.)
Z xZ y Z 1
1 f(y)
Define T : C[0, 1] → C[0, 1] by (T f)(x) = 2π
dt dy − dy. Since C[0, 1] is a
0 0 2 + t 0 2 + (xy)π
complete metric space with d(f, g) = kf − gk∞ and
Z 1 Z 1
f(y) g(y) 1
|T f(x) − T g(x)| = dy − dy ≤ kf − gk∞ .
0 2 + (xy)π 0 2 + (xy)π 2

By the contractive mapping theorem, there exists f ∈ C[0, 1] such that T f = f and we are done.

4. (For a binomial coefficient problem, we should think about the binomial expansion of (1 + x)n .) Observe
X p    X p   
p p+j p p+j
the sum I = = is the coefficient of xp in the expansion of
j j j p
j=0 j=0

p  
X p  
X 
p p
(1 + x)p+j = (1 + x)j (1 + x)p = (2 + x)p (1 + x)p .
j j
j=0 j=0

1
 X p     
p p p p k p
Expanding (2 + x) (1 + x) , we get I = 2 . Since p divides for 0 < k < p, we have
k p−k k
k=0
     
p p 0 p p p
I≡ 2 + 2 = 2p + 1(mod p2).
0 p p 0
1 cos(t ln n)  1 2 1 1
5. Note Re 1+it = Re e− ln n−it ln n = and x + x2 = x + − ≥ − . Let w = t ln 2. Then
n n 2 4 4
cos w cos(t ln 3) cos 2w cos(t ln 5)
Re h(1 + it) = 1 + + + +
2 3 4 5
2
cos w 1 2 cos w − 1 1
≥ 1+ − + −
2 3 4 5
1 1 1 cos w + cos2 w 13 1
= 1− − − + ≥ − > 0.
3 4 5 2 60 8
6. There are two solutions to this algebra problems. The first one is by linear algebra techniques. The
second one is by field theory techniques.
(Solution 1 due to Lau Lap Ming) Recall n = [K : F ] is the dimension of K as a vector space over
F. Let v1, v2 , · · ·, vn be a basis of K over F. We claim it is also a basis of K(ζ) over F (ζ). Then
[K(ζ) : F (ζ)] = n = [K : F ].
Suppose there are c1, c2, . . . , cn ∈ F (ζ) such that c1v1 + c2v2 + · · · + cn vn = 0. Now each ci is of the
form pi (ζ)/qi (ζ), where pi and qi are polynomials with coefficients in F and qi(ζ) 6= 0. By taking common
denominators, we may assume all qi(ζ) are equal, say to q(ζ). Let pi(ζ) = ai0 + ai1ζ + · · · + aim ζ m ,
where aij ∈ F and m is the maximum degree of p1, p2, . . . , pn. Then
X n n X
X m  m X
X n 
0 = q(ζ) ci vi = aij ζ j vi = aij vi ζ j .
i=1 i=1 j=0 j=0 i=1
X
n
Since ζ is not a root of nonconstant polynomials over K, we have all aij vi = 0 for j = 0, 1, . . ., m.
i=1
By the linear independence of vi in K, we get all aij = 0, which imply all ci = 0. So v1, v2 , · · ·, vn are
linearly independent in K(ζ).
Next, if b0 +b1 ζ +· · ·+bk ζ k ∈ K(ζ) with bi ∈ K = span{v1, v2 , . . ., vn}, then write each bi = βi1 v1 +
k X
X n n X
X k 
βi2 v2 + · · ·+ βin vn , where βij ∈ F. Then b0 + b1ζ + · · · + bk ζ k = βij vj ζ i = βij ζ i vj is in
i=0 j=1 j=1 i=0
the span of v1 , v2, . . . , vn over F (ζ). The claim is proved.

(Solution 2) Since K is obtained by adjoining finitely many algebraic elements to F, inductively, we may
suppose K = F (α) for some algebraic α ∈ C over F. Let
f(x) = xn + an−1xn−1 + · · · + a0 ∈ F [x]
be the minimal polynomial of α over F. Clearly, f(x) ∈ F (ζ)[x] annihilates α. It is enough to show
f(x) ∈ F (ζ)[x] is also the minimal polynomial of α over F (ζ) because then
[K(ζ) : F (ζ)] = [F (ζ)(α) : F (ζ)] = n = [K : F ].
(∗)
Suppose g(x) = xm + gm−1(ζ)xm−1 + · · · + g0(ζ) ∈ F (ζ)[x] is another polynomial such that g(α) = 0,
where gi (ζ) ∈ F (ζ). Since F ⊆ C, F is an infinite field, one can find u ∈ F such that the product
p(ζ) of the denominators of gi(ζ) ∈ F (ζ) do not annihilate u when ζ is replaced by u. Since p(ζ)(αm +
gm−1 (ζ)αm−1 + · · · + g0(ζ)) = 0, the polynomial p(x)(αm + gm−1 (x)αm−1 + · · · + g0 (x)) is the zero
polynomial in F (α)[x] = K[x]. Since p(u) 6= 0, we get
αm + gm−1 (u)αm−1 + · · · + g0 (u) = 0,
where gi (u) ∈ F. Since f is the minimal polynomial of α over F, this implies m ≥ n. Therefore,
f(x) ∈ F (ζ)[x] is the minimal polynomial of α over F (ζ).