solutiones economía

8/10/2019 Solutiones Economa

1/6

Mathematics for Economics

Pierpaolo De Blasi

e-mail: [email protected]

I.3. Vector spaces, linear independence and basis

1. Show that the following subsets ofR2 are subspaces:

a) {(x, y) : x= y}

b) {(x, y) : x+ 4y= 0}

Solution. Remember that a subset W of R2 is a subspace if (i) v+ w W for anyv, w W; (ii)cv W for any v W and c R; (iii) 0= (0, 0) W.

a) An element of W is of the form (a, a) for a R. Let a1, a2 be two arbitrary real

numbers and set v = (a1, a1) and w = (a2, a2). (i) v+ w = (a1+ a2, a1+ a2) Wsince a1+ a2 R. (ii) Let c R. Then cv = (ca1, ca1) W. (iii) The zero vector0= (0, 0) clearly belongs to W.

b) An element of W is of the form (4a, a) for a R. Let v = (4a1, a1) and w =(4a2, a2) fora1, a2 R. (i) v +w= (4a14a2, a1+a2) = (4(a1+a2), (a1, a2)) W(ii) Let c R. Then cv = (c(4a1), ca1) = (4(ca1), ca1) W. (iii) Since the zerovector can be written as 0= (4(0), 0), it belongs to W.Alternatively, use Example 1.3.6 in the lecture notes for a= (1, 4).

2. Let U and W be subspaces ofRn. Denote by U W (the intersection ofU andW) the set of elements which lie both in U and Wand by U+ W ={v V : v=u + w, u U, w W}.Show that a) U Wand b) U+ W are subspaces.

Solution.

a) Let v1, v2 U W, that is

v1 U, v1 W and v2 U, v2 W

(i) Since both v1+ v2 U and v1+ v2 W(they both are subspaces ofV), v1+ v2U W. (ii) For c R, both cv1U and cv1W, thereforecv1 U W. (iii) Sinceboth 0 U and 0 W, 0 U W.

1


2/6

b) Let v1, v2 V, u1, u2 U and w1, w2 W such that

v1= u1+ w1, v2= u2+ w2

that is two arbitrary elements ofU+W. (i) Write v1+ v2= (u1+ u2)+(w1+w2) Sinceu1+ u2 Uand w1+ w2 W, v1+ v2 U+ W. (ii) Forc R, writecv1= cu1+cw1.Since cu1U and cw1W, cv1 U+W. (iii) 0= 0 + 0 and 0 U, 0 W, hence0 U+ W.

3. a) Let Vbe a subspace ofRn. Let V be the orthogonal complement ofV, thatis the set of elements ofRn which are orthogonal to every elements ofV. Showthat V is a subspace ofRn.

b) Let V = L(v1, . . . , vr), i.e. the subspace generated by v1, . . . , vr and let Wbe the orthogonal complement of {v1, . . . , vr} (set of x R

n such that x isorthogonal to each vi, i = 1, . . . , r). Show that W =V

.

Solution.

a) Let w1, w2 be two arbitrary elements ofV, that is for any v V,

w1 v= 0 and w2 v= 0

(i) w1+ w2 V since (w1+ w2) v= w1 v + w2 v= 0 + 0 = 0 for any v V. (ii)

For c R, we have (cw1) v= c(w1 v) =c(0) = 0, for any v V, hence cw1 V.

(iii) Clearly 0

V

since 0

v

= 0 for any v

Vb) Since w V is orthogonal to any vi,i = 1, . . . , r, clearlyV

W. We show next theconverse, W V, that is any w W is orthogonal to any v V. Let w W, that is

w vi = 0, i= 1, . . . , r

SinceV is generated by v1, . . . , vr, any v Vcan be written as linear combination ofv1, . . . , vr, i.e. there exist a1, . . . , ar Rsuch that v= a1v1+. . .+arvr Now

w v=r

i=1

ai(w vi) =r

i=1

ai(0) = 0

as was to be proved.

4. LetS,S1and S2be convex subsets ofRn. Show that the following sets are convex:

a) {v Rn : v= c s, s S} for a fixed c R.

b) S1+S2

Solution. Let t be an arbitrary number between 0 and 1. Remember that a set S isconvex if (1 t)s1+ts2 Sfor any pair s1, s2 Sand any t [0, 1].

2


3/6

a) Take two arbitrary w1, w2 {v Rn : v= c s, s S}, that is w1= c s1and w2= c s2

for some s1, s2 S. Then

(1 t)(c s1) +t(c s2) =c[(1 t)s1+ts2) S

]

which belongs to{v Rn : v= c s, s S}.

b) Let v, w S1+ S2, that is v = v1+ v2 for some v1, v2 S1 and w = w1+ w2 forsome w1, w2 S1. For any t [0, 1] we have

(1 t)v +tw= (1 t)(v1+ v2) +t(w1+ w2)

= [(1 t)v1+tw1

S1] + [(1 t)v2+tw2

S2]

that is (1 t)v +tw S1+S2 as was to be proved.

5. Express the given vector v as a linear combination of the given vectors a, b.

a) v= (1, 0), a= (1, 1), b= (0, 1)

b) v= (2, 1), a= (1, 1), b= (1, 1)

c) v= (1, 1), a= (2, 1), b= (1, 0)

d) v= (4, 3), a= (2, 1), b= (1, 0)

Solution. We look for the solution to the system

Ax= v, where x= (x, y), A= [a b]

corresponding to the linear relationx a +y b= v.

a) x = 1x+y = 0

E12(1)

x = 1

y = 1

b) x+y = 2

x+y = 1E12(1)

x+y = 2

2y = 3E2(1/2)

x+y = 2

y = 3/2E21(1)

x = 1/2

y = 3/2

c) x = 1

y = 1

d) x = 3

y = 2

3


4/6

6. Let a1, . . . , ar be nonzero vectors which are mutually orthogonal (ai aj = 0 when

i=j). Show that they are linearly independent.

Solution. Let x1, . . . , xn be real numbers such that

x1a1+. . .+xrar = 0

Then we need to show that xi = 0 for anyi = 1, . . . , rin order to prove linear independence.For arbitrary i, take the scalar product of both sides with ai to get

x1(a1 ai) +. . .+xr(ar ai) = 0 ai; xi(ai ai) = 0

since ai aj = 0 when i=j . Since ai ai = ai2 >0 (ai = 0 by assumption), it must be

thatxi = 0. Given the arbitrariness ofi, the proof is complete.

7. In each of the following cases find a basis for the given spaces and determine thecorresponding dimension.

a) space of 2 2 matrices

b) space ofm n matrices

c) space ofn n diagonal matrices

Solution. The general strategy is to put the elements in one-to-one (bijective) relationwith Rn for suitable n and determine a basis in correspondence of a basis ofRn, so thatthe dimension isn.

a) Possible basis is 1 00 0

,

0 10 0

,

0 01 0

,

0 00 1

so that dimension is 4.

b) Every (m n) matrixA is in one-to-one relation with the mn-dimensional space Rmn,by stacking the columns ofA from the first to the last, for example,

a11a12a21a22

a31a32

a11a21a31a12a22a32

Since a basis ofRmn is given by the mn standard unit vectors e1, . . . , emn, translatingthem back into matrix form by the reverse operation, we obtain a basis for the spaceof (m n) matrices:

Iij , i= 1, . . . , m;j = 1, . . . , n

where Iij is the zero matrix with 1 into the (i, j) position. In fact A = (aij) can bewritten as the linear combination

A=i,j

aijIij

The dimension is then mn.

4


5/6

c) Since there is a one-to-one relation between a diagonal matrixA with diagonal element

a1, . . . , an and the vector a= (a1, . . . , an) Rn

, reasoning as before a basis is given byIii fori= 1, . . . , nand the dimension is n.

8. LetWbe a subspace ofR3. What are the possible dimensions for W ifW = R3?

Solution. Since the dimension ofR3 is 3, the dimension of any vector subspace W= R3

is


6/6

a) In order to show thatW, the space of polynomials of order n, is not a vector space it is

sufficient to show that it is not closed with respect to the sum. In fact, ifp1(x) = a1 xis a polynomial with coefficient vector a1 = (a10, . . . , a1n) and a1n = 0, the sum withany other polynomial p2(x) = a2 x with a2n = a1n gives

p1(x) +p2(x) = b x=ni=0

(a1i+a2i)xi

wherebn = a1n+a2n = 0, that is a polynomial of order < n.

Alternatively, we may considerWa subset of the space of all real-valued function whosezero element is the function identically equal to zero. Then it is easy to see that Wis not a vector subspace because the zero function (polynomial with a = 0) is not apolynomial of degreen.

b) If we now denote by Wn the set of polynomials of order n, the set of polynomials oforder n is W =

ni=0Wi. In particular,p W is p(x) = a xwhere a is free to vary

in Rn+1 (no constraint on an). ThereforeW is in one-to-one relation with the vectorspace Rn+1.

Alternatively, one could prove that Wsatisfies (i)-(ii)-(iii), hence is a vector subspaceof the space of real-valued functions.

6

solutiones economía

Documents