Linear Algebra & Geometry: solutions 4
1. We consider v+w
2
= (v + w) (v + w) = v v + 2v w + w w = v 2 + w 2 + 2v w w and obtain w
Now we use Cauchy Schwarz v w |v w| v v+w
2
v 2+ w 2+2 v = ( v + w )2
and taking the square root gives the triangle inequal
Linear Algebra & Geometry: Solutions to sheet 8
1. (a) Let x Rn , then R T (x) = R(T (x), S T (x) = S (T (x) and (R + S ) T (x) = (R + S )(T (x) = R(T (x) + S (T (x). (b) Let x Rn , then (R S ) T (x) = R S (T (x) = R(S (T (x) and R (S T )(x) = R(S T (x) =
Linear Algebra & Geometry: Solutions to Sheet 7
1. We have to check that (i) T (x + y) = T (x) + T (y) and (ii) T (x) = T (x). (a) T (x, y ) = (x y, 5x) is linear. (i) let x = (x, y ), y = (x , y ), then T (x + x , y + y ) = (x + x y y , 5x +5x ) = (x y +
Linear Algebra & Geometry: Sheet 6
Set on Friday, November 12: Questions 1, 2, and 3
1. Let V Rn be a linear subspace and u1 , u2 , , uk an orthonormal basis of V . (a) For x Rn we say that x V if for any v V we have x v = 0. Show that x V is equivalent t
Linear Algebra & Geometry: Solutions to sheet 5
1. We have to show that if
k
i vi = 0
i=1
then 1 = 2 = = k = 0. Let us take the dot product of the sum with vj , then since vj vi = 0 if i = j we nd
k k
vj
i=1
i vi
=
i=1
i vj vi = i vj vj ,
but vj = 0 and
Linear Algebra & Geometry: Sheet 4
Set on Friday, October 30: Questions 2, 3, 4, and 6
1. Use the Cauchy Schwarz inequality |v w| v w to derive the triangle inequality
v+w v + w Hint: express v + w
2
in terms of the dot product.
2. Use the Cauchy Schwarz
Linear Algebra & Geometry: Sheet 8
Set on Friday, November 27: Questions 1, 2, 3 and 5
1. Show that (a) if T : Rn Rm and R, S : Rm Rk are linear maps, then (R + S ) T = R T + S T (b) if T : Rn Rm , S : Rm Rl and R : Rl Rk are linear maps then (R S ) T = R
Linear Algebra & Geometry: Sheet 3
Set on Friday, October 23: Questions 1, 2, 3,4
1. Use the relation ei = cos + i sin to prove De Moivre s Theorem : for any n N cos(n) + i sin(n) = (cos + i sin )n Use this formula to derive the following relations cos(3)
Linear Algebra & Geometry: Sheet 7
Set on Friday, November 20: Questions 1, 2, and 3
1. Determine which of the following maps are linear maps (a) T : R2 R2 , T (x, y ) = (x y, 5x). (b) T : R2 R2 , T (x, y ) = (x y 2 , 5x). (c) T : R2 R2 , T (x, y ) = (x 1
Linear Algebra & Geometry: Sheet 2
Set on Friday, October 16: Questions 1, 2, 4 (a)(c)(e), 5 (i), 6
1. Consider the following vectors: ( a) 1 1 (b) 0 5 cos sin (c) 3 4 ,
nd for each of them a > 0 and a [0, 2 ) such that v = u() where u() = .
2. Find the c
Linear Algebra & Geometry: Sheet 5
Set on Friday, November 6: Questions 1, 2, 3, and 4
1. Show that if the k vectors v1 , v2 vk Rn are all non-zero and mutually orthogonal, i.e., vi vj = 0 if i = j , then they are linearly independent. 2. Recall the denit
Linear Algebra & Geometry: Sheet 9
Set on Friday, Dec 4: Questions 1, 2, 3 and 4
1. Consider the following matrices A= 00 12 B= 1 1 C= 4 2 0 1 D = 2 0 31 E= 2 01 1 3 0
and determine which of them can be multiplied, and in which order, and compute the prod
ANALYSIS EXERCISE 15SOLUTIONS
1. Set Am := 1 an , A := m 1 ( an + bn ). Then
m
1
an , Bm :=
m 1 bn ,
B :=
1 bn
and Cm :=
Cm = A m + B m . Also, Am A as m and Bm B as m from the denition of convergence. By properties of convergent sequences (Theorem 1.1.
ANALYSIS EXERCISE 13SOLUTIONS
1. For the function f (t) = tp write the Mean value theorem f (x) f (y ) = f ( )(x y ), where x > > y > 0, i.e. xp y p = p p1 (x y ). It is enough to notice that y p1 < p1 < xp1 . 2. (a) xn 1 nxn1 = lim = n. x1 x2 x x1 2x 1 l
ANALYSIS EXERCISE 12SOLUTIONS
1. Since the function f is continuous on [a, b] there are cfw_x1 , x2 [a, b] such that f (x1 ) = m, f (x2 ) = M . If m = M then f is a constant function, and the result is vacuously true (as (m, M ) = .) Otherwise set x = mi
ANALYSIS EXERCISE 11SOLUTIONS
1. By the denition of continuity for every x0 [a, b]
xx0
lim f (x) = f (x0 ).
Using the Heine denition one can compute the above limit along the sequences xn x0 , xn = x0 . Choose xn Q. Then f (xn ) = 0 for all n N, and it is
ANALYSIS EXERCISE 9SOLUTIONS
1. (a) Let lim f (x) = A. By the denition
x a
( > 0)( > 0)(x D(f ) (0 < |x a| < ) (|f (x) A| < ) . Denote h = x a, then x = a + h. We obtain ( > 0)( > 0)(a + h D(f ) (0 < |h| < ) (|f (a + h) A| < ) . (b) Write down the denitio
ANALYSIS EXERCISE 8SOLUTIONS 1. First, using the inequality 1 (a + b) ab, which is valid for all 2 non-negative a and b we see that an p for all n 2. So the sequence is bounded below. Next, we prove that it is decreasing. Indeed, an+1 an = 1 2 p an an = p
ANALYSIS EXERCISE 7SOLUTIONS
1. (a) Since (n N+ ) (an = a), one has evidently that ( > 0)(N N)(n N)[(n > N ) (|an a| = 0 < )] (Just take N = 1). (b) By the denition (M > 0)(N N)(n N) [(n > N ) (an > M )]. Note that (an > M ) ( a1 < n the above. Then
1 M )
ANALYSIS EXERCISE 6SOLUTIONS
1. 1 (a 1)2 2= 0. a a The equality holds i a = 1. a+ 2. Multiplying the inequalities a + b 2 ab, a + c 2 ac, c + b 2 cb, we obtain the desired. The equality holds i a = b = c. 3. For all n 2 we have 1 1 1 1 < = . 2 n n(n 1) n1
ANALYSIS EXERCISE 5SOLUTIONS 1. (a) Base case : n = 10; 210 = 1024 > 1000 = 103 . The statement is true. Induction step: Suppose that 2k > k 3 (k 10). We have to prove that 2k+1 > (k + 1)3 . By the assumption 2k+1 = 2 2k > 2k 3 . So it is sucient to prove
ANALYSIS EXERCISE 4SOLUTIONS
1. (a) Let x1 , x2 X be arbitrary. Suppose that g (f (x1 ) = g (f (x2 ). Then f (x1 ) = f (x2 ) since g is injective. Therefore x1 = x2 since f is injective. This proves that g f is injective. (b) Let z Z be arbitrary. Then y
ANALYSIS EXERCISE 3SOLUTIONS
1. [in F.T.A.] (a) Reexive (f (x) = f (x), symmetric (f (x) = f (y ) (f (y ) = f (x), transitive ([(f (x) = f (y )] [f (y ) = f (z )] [f (x) = f (z )]). (b) Dene f : R2 R by f (x, y ) = x2 + y 2 . Then [(a, b) (c, d)] [f (a, b
ANALYSIS EXERCISE 2SOLUTIONS
1. (a) (x (Ac )c ) (x Ac ) (x A) (x A). (b) (x (A B )c ) (x (A B ) (x A) (x B ) [(x A) (x B )] [(x Ac ) (x B c )] (x Ac B c ). (c) (x (A B )c ) (x (A B ) (x A) (x B ) [(x A) (x B )] [(x Ac ) (x B c )] (x Ac B c ). 2. (a) (x A