Review Problems ICME ad MS&E Refresher Course September 9, 0 Warm-up problems. For the followig matrices A = 0 B = C = AB = 0 fid all powers A,A 3,(which is A times A),... ad B,B 3,... ad C,C 3,... Solutio: It is easy to verify that (a) (b) (c) B = B 4 = = A = A = A 3... [ ] 0 0 C = C 3 = = B = B 3 = B 5 =... [ ] 0 0 0 0. Use Ve diagram to prove the followig distributive law of sets: A (B C) = (A B) (A C). Solutio: For a good illustrative solutio, please visit the followig website: http://pirate.shu.edu/~wachsmut/ira/logic/proofs/distlaw.html 3. Which of the followig fuctios are ijective (oe-oe), surjective (oto), or bijective (oe-oe ad oto)? The domai for all fuctios is the etire real lie R: f(x) = x, g(x) = cos(x), h(x) = x 4. 4. Let A ad B be two sets. We say that A ad B have the same cardiality if there is a bijective fuctio f from A to B. We deote this as card(a) = card(b). If there exists a fuctio f from A to B that is ijective, what ca you say about card(a) ad card(b). What if f is surjective? Solutio: If f is surjective, card(a) card(b), if f is surjective the iequality is reversed. 5. A sequece {x } = is called mootoe icreasig if x + x for all. A mootoe decreasig sequece is defied similarly. Commet whether the followig sequeces are mootoe icreasig or decreasig: { }, = Solutio: { }. + =
(a) (b) x = x + + therefore, x x + ad this sequece is mootoic decreasig. x x + = + + + = (+) 0 therefore, x x + ad the sequece is mootoic icreasig. 6. Deviatio of middle elemet value from average: Suppose x is a -vector, with = m ad m. We defie the middle elemet value of x as x m. Defie: f(x) = x m x i, which is the differece betwee the middle elemet value ad the average of the coefficiets i x. Express f i the form f(x) = a T x, where a is a -dimesioal vector. This meas f is a liear fuctio of x. Solutio: It is easy to verify that implies that f(x) = x m i= i= x i = x x +( )x m x { a i = if i m ( ) if i = m 7. Permutatio matrices: A square matrix A is called a permutatio matrix if it satisfies the followig three properties: all elemets of A are either zero or oe each colum of A cotais exactly oe elemet equal to oe each row of A cotais exactly oe elemet equal to oe. For example: 0 0 A = 0 0 0 0 is a permutatio matrix. I other words, a permutatio matrix is the idetity matrix with its rows reordered. (a) Let A be a permutatio matrix. Give a simple descriptio i words of the relatio betwee a vector x ad f(x) = Ax. Solutio: Give a permutatio of elemets we may write the permutatio matrix A as π : {,...,} {,...,} A π = [ e π() e π()... e π() ] where, e i deotes a -vector with all etries zero except the i-th etry. Multiplyig the matrix A π by a -vector x results i swappig the i-th elemet with the π(i)-th elemet. I other words f(x) i = x π(i)
(b) We ca also defie a secod liear fuctio g(x) = A T x, i terms of the same permutatio matrix. What is the relatio betwee g ad f? Hit: Cosider the compositio g(f(x)). First try a few simple examples with = ad = 3, ad the geeralize. Solutio: It ca be show that the permutatio matrices are orthogoal, so A T = A ad A T A = I. For a proof, visit http://e.wikipedia.org/wiki/permutatio_matrix So, Slightly harder problems g(f(x)) = A T f(x) = A T Ax = x. Recall the Cauchy-Schwarz iequality for two dimesioal vectors x ad y: x T y x y. Use this to show that the arithmetic mea of a set of umbers (give by a vector x) is greater tha or equal to the harmoic mea. That is, show that: x i i= ( x i= i ). whe, x i 0,i =,..., Solutio: We write the Cauchy-Schwartz iequality i terms of a, b a T b a b. Now apply the Cauchy-Schwartz iequality with a i = x i,b i = xi, so that ( a T b = a = ( x i ) / b = i= x i= i Pluggig it ito the Cauchy-Schwartz iequality, squarig ad rearragig, we obtai our desired solutio.. A orm is a fuctio from f : A R with the followig properties: (a) f(x) 0 for all x A ad f(x) = 0 if ad oly if x = 0. (b) f(αx) = α f(x), for α R. (c) f(x+y) f(x)+f(y) for x,y A. Show the followig are orms: ) / (a) For A = C ([0,]) (that is the set of all cotiuous fuctios o [0,]), show that is a orm. f = sup f(x) x [0,] Solutio: (a) Observe that f = sup f(x) f(0) 0 x [0,] 3
thus the orm is o-egative. Further For (b) we see that f(x) = 0 f(x) = 0 sup f(x) = f(x) = 0. x [0,] αf = sup αf(x) = sup α f(x) = α f. x [0,] x [0,] Ad fially, we have (c) via the triagle iequality, f +g = sup f(x)+g(x) sup ( f(x) + g(x) ) sup f(x) + sup g(x) = f + g. x [0,] x [0,] x [0,] x [0,] (b) For A = R, show that with x = {ν i : i } R is a orm. Solutio: Agai for (a) observe that ad that For (b) observe that x = x = ν i i= ν i ν 0 i= x = 0 ν i = 0 i ν i = 0 i αx = αν i = i= i= ν i = x = 0. i= α ν i = α ν i = α x. Part (c) follows from the triagle iequality (let y = {ξ i : i } R ): x+y = ν i +ξ i i= ν i + ξ i = 3. The iduced matrix orm for a give vector orm is defied as i= i= ν i + i= A = sup Ax. x = ξ i = x + y. i= For the vector orm determie the iduced matrix orm. x = max i ν i Solutio: Assume A R m. We start with the defiitio: A = sup Ax x R : x = = sup max x R : x = i m α ij ν j j= 4
We ow boud α ij ν j by j α ij ν j, which yields sup max x R : x = i m j= α ij ν j Ad ow we boud j= α ij ν j by (max j ν j ) j= α ij, which gives sup max x = i m As x =, we have that max j ν j so this becomes which is ow idepedet of x so that sup max x = = max i m j= i m j= α ij. ( ) max ν j α ij j α ij Thus we have a upper boud for A. If we ca fid a vector i {x R : x = } such that Ax achieves this boud, the we are doe. j= Suppose that the Ith row is has the maximal row sum, so that max i m j= α ij = α Ij. j= Defie a vector y = {ξ i : i } as: ξ i = { α ij 0 α ij < 0 Observe that y =. Further more, by costructio, if Ay = b = {β i : i m}, β I β i for all other rows. Thus Ay = max i m j= α ijξ j = j= α Ij = max i m j= α ij which is the upper boud. 4. I a fiite dimesioal space, e.g. R, all orms are equivalet i the sese that there are costats m ad M that are fuctios of the dimesio of the space () such that M x α x β m x α. Fie m ad M whe α = ad β =. (This allows us to prove theoretical covergece i a coveiet orm ad check it - umerically - i aother). Solutio: We start with fidig M : x M x. Defie a vector ˆx = { ν i : i }, the x = ν i, i= 5
ad we ow use the defiitio of ˆx to write this orm as a dot product ad add the absolute value, as the absolute value of a positive umber does t chage aythig ˆx T e = ˆx T e. We ow apply the Chauchy-Schwartz Iequality which states that a T b a b, so that ad ote that ˆx = x, so ˆx e = x. To prove that this upper boud is tight, set x = e ad observe that e = ad that e =. Thus M =. To obtai the lower boud observe that ( x = ν i ) = ν i ν j. i= i= j= We ow group terms that are squared ad the others so that, = ν i + ν i ν j i= i= j i ad drop the secod term (as it oly icreases the sum) so that ν i = x. Thus x x. To prove the boud is tight take x = e i. Thus m =. i= 6