Math 6CM - Solutios to homework 3 Cédric De Groote October 2 th, 208 Problem : Let F be a field, m 0 a fixed oegative iteger ad let V = {a 0 + a x + + a m x m a 0,, a m F} be the vector space cosistig of all polyomials over F of degree at most m. Suppose that p,, p m+ V are polyomials such that p j () = 0 for all j. Prove that the vectors p,, p m+ V are liearly depedet. Solutio: Notice that V is a vector space of dimesio m + ; a explicit basis is {, x, x 2,, x m }. Cosider the liear trasformatio T : V F : p p(). This is ideed liear: for ay λ, µ F ad p, q V, we have T (λp+µq) = (λp+µa)() = λp()+µq() = λt (p) + µt (q). Notice that Ra(T ) = F: for ay a F, the costat polyomial p(x) = a satisfies p() = a. Therefore, the rak of T is. It follows from the rak-ullity theorem (Theorem 7.7 i the otes) that the dimesio of the kerel of T is dim V dim Ra(T ) = (m + ) = m. By defiitio of p,, p m+, they all belog to the kerel of T. So, these are m + vectors i a subspace of V which has dimesio m. These ca ot be liearly idepedet. To see that, we ca for example use the Theorem 5.9 i the otes: if they were liearly idepedet, we could extet this collectio of vectors ito a basis of the kerel. So this basis would have at least m + elemets, but the dimesio of the kerel of T is m ad all the bases have the same umber of elemets, which is a cotradictio. Problem 2: Let V be a fiite dimesioal vector space over a field F ad suppose that U,, U m are subspaces of V. Defie U + + U m = {u + + u m u j U j for all j m}. Prove that dim(u + + U m ) dim U + dim U m. Solutio: Let A i be a set of vectors of V that form a basis of U i ; say that the cardiality of A i is k i (so that U i has dimesio k i ). The, the set B 0 of k +k 2 + +k m vectors obtaied by takig together all the vectors of all the A i s (maybe there are repetitios, but that s fie) geerates U + + U m : ideed, u + + u m is a liear combiatio of the u k s, ad each u i is a liear combiatio of elemets of A i, so u + + u m is a liear combiatio of elemets of B 0.
If B 0 cosists of liearly idepedet vectors, the it is a basis of U + + U m, which therefore has dimesio k + + k m. I this case, we have so we are good. dim(u + + U m ) = dim U + dim U m, If B 0 is liearly depedet, the by the problem 4 of homework 2, there is a set B B 0 with oe less elemet that still spas U + + U m. If B is liearly idepedet, the we have proved that B is a basis of U + + U m, ad so we have so we are good. dim(u + + U m ) = dim U + dim U m, We ca keep goig this way: fid a sequece of set B 2 B B 0, with B i+ havig oe less elemets tha B i. If ay of these is liearly idepedet, we are doe: if B i is liearly idepedet, the we have proved that dim(u + + U m ) = dim U + dim U m i, so we are good. If they are all liearly depedet, the we ca keep goig. Sice U,, U m are subspaces of a fiite dimesio vector space, they must have fiite dimesio, hece B is fiite. So this process must create at some poit some B i cosistig of oly oe vector. Uless that vector is 0, it must be liearly idepedet, which cotradicts the fact that we assumed all the B i s to be liearly depedet. Fially, otice that it is ot possible that this fial vector is 0, as it was a elemet of a basis of oe of the U i s. Remark 0. Everythig after the first paragraph i this proof basically proves that if we have a set of vectors that geerates a (fiite dimesioal) vector space, the there exists always a subset of that set which is a basis of the vector space. Compare with Theorem 5.4 i the otes, which says that a set of liearly idepedet vectors i a vector space ca always be exteded to a basis of the vector space. Problem 3: Determie all values of λ R such that the collectio of vectors i R 3 is liearly depedet. (λ,, ), (, λ, ), (,, λ) Solutio: I will ot write all the details here; we will see later that the otio of determiat is very helpful to tackle this kid of problem: the determiat of the matrix λ λ λ is zero oly whe λ = or λ = 2, so these are the oly values of λ for which these vectors are liearly depedet. 2
Sice we do t kow about determiats yet, here is how to proceed: assume that there exist a, b, c R such that a(λ,, ) + b(, λ, ) + c(,, λ) = (0, 0, 0). This is equivalet to the system i a, b, c: λa + b + c = 0 a + λb + c = 0 a + b + λc = 0. Let s try to perform Gauss elimiatio i this matrix. Whe doig so, you multiply/divide rows by certai expressios i terms of λ, ad substract/add rows to each other. You ca oly do so if these expressios are all ozero. Assumig that these expressios are all ozero, after you are doe with Gauss elimiatio, you fid yourself with a diagoal matrix, whose diagoal etries are all ozero. The oly solutio to that system is a = b = c = 0, which proves that (λ,, ), (, λ, ), (,, λ) are liearly idepedet for all values of λ so that these expressios are ozero. Fially, there is a fial set of λ s that make some of these expressios equal to 0. For these λ, you ca check that the correspodig (λ,, ), (, λ, ), (,, λ) are liearly (i)depedet, by just tryig each λ separately. You should get to the same aswer as above: λ = ad λ = 2 make this three vectors liearly depedet, ad all the other values of λ that make oe of these expressios be equal to 0 produce three vectors which are liearly idepedet. Ayway, this is all log, complicated ad aoyig to write dow. As I said above, this kid of exercise will be much faster oce we will kow about determiats. Problem 4: Suppose that V is a vector space over F, that v,, v k V are liearly idepedet, ad that u V is ay other vector. Show that Ca this dimesio be equal to k? dim spa (v + u,, v k + u) k. Solutio: Notice that v j+ v j = (v j+ + u) (v j + u), hece v j+ v j spa (v + u,, v k + u) for 2 j k. We claim that these k vectors are liearly idepedet. By the Theorem 5.9 i the otes, this will imply that there exist a basis of spa (v + u,, v k + u) cotaiig these k vectors, hece dim spa (v + u,, v k + u) k. Now let s see that they are liearly idepedet. Assume that there exists a,, a k F such that a (v 2 v ) + + a k (v k v k ) = 0. Rearragig, we get ( a )v + (a a 2 )v 2 + + (a k 2 a k )v k + a k v k = 0. Sice v,, v k are liearly idepedet, this meas that all the coefficiets are 0, which meas that all the a i s are zero. Hece the vectors v 2 v,, v k v k are liearly idepedet, ad we are doe. 3
This dimesio ca be equal to k : take V = F k, ad the v i s formig the stadard basis. Take u = v. The dim spa (v + u,, v k + u) = dim spa (0, v 2 v,, v k v ) = dim spa (v 2 v,, v k v ), sice the vector 0 ever cotributes to a spa. This last set has k vectors, hece its dimesio is at most k. Sice we kow by the above that it is also at least k, it must be equal to k. Problem 5: Let p be a prime ad F p be the fiite field with p elemets. subspace of F p of dimesio d has exactly p d elemets. Prove that every Solutio: Let W be a subspace of F p of dimesio d, ad let v,, v d form a basis of it. The, for each w W, there exist c,, c d F p such that w = c v + + c d v d. Furthermore, this collectio of umbers is uique: if c v + + c d v d = w = c v + + c d v d, the (c c )v + + (c d c d )v d = 0, hece c i = c i for each i d, sice the vectors v,, v d are liearly idepedet. This meas that to each vector w W, we ca associate a uique d-uple (c,, c d ) of umbers i F p. So, there are as may vectors i W as d-uples of umbers i F p. Ad clearly, there are p d such d-uples. Remark 0.2 The umbers c,, c d above are called the coordiates associated to the basis v,, v d of W. Problem 6: We say that x is a fixed poit of a fuctio f if f(x) = x. Let f(x) be a cotiuous fuctio o [0, ] such that 0 f(x) for all x [0, ]. (a) Show that there exists x [0, ] such that f(x) = x. (b) Let f(x) be a odecreasig fuctio o [0, ] such that 0 f(x) for all x [0, ]. Show that for ay x [0, ], either x is a fixed poit of f or the sequece y, defied recursively as y = x ad y + = f(y ) coverges to a fixed poit of f. Solutio: (a) Cosider the fuctio g(x) = f(x) x. Notice that, by the assumptios o f, we have g(0) = f(0) 0 0 ad g() = f() 0. If ay of these two iequalities is a equality, we are doe. If these two iequalities are strict, the by the itermediate value theorem (g(0) > 0 ad g() < 0), there must exist x [0, ] such that g(x) = 0. This traslates to f(x) = x. (b) If f(x) = x, the we are doe. Otherwise, assume that f(x) > x, i.e. y 2 > y. The, because f is odecreasig, we have f(y 2 ) f(y ), i.e. y 3 y 2. A easy iductio the shows that the sequece y is odecreasig. Coversely, if f(x) < x, the y 2 < y, ad applyig f we get f(y 2 ) f(y ), which is y 3 y 2. Agai, a easy iductio shows that y is oicreasig. Attetio! f beig odecreasig does ot mea that x f(x). It meas that if a < b, the f(a) f(b). 4
Furthermore, sice f is bouded above by, this sequece is bouded above by. By the Theorem 3. i the otes, it must coverge to some y [0, ]. We show that y is a fixed poit of f. Suppose that f(y) y, ad let ε = f(y) y 3 > 0. By cotiuity of f, there must exist δ > 0 such that if x y < δ, the f(x) f(y) < ε. Sice y coverges to y, there exists N N such that y y < mi(δ, ε) for N. By defiitio of δ ad cotiuity of f, we must therefore have f(y N ) f(y) < ε. But the we have a cotradictio. f(y) y f(y) f(y N ) + f(y N ) y = f(y) f(y N ) + y N+ y = ε + ε 2 f(y) y, 3 Problem 7: (a) Let f(x) be a cotiuous fuctio o [0, ] such that 0 f(x) for all x [0, ], ad f(0) = f(), ad let N. Show that there exist 0 x < x 2 such that f(x ) = f(x 2 ) ad x 2 = x + /. (b) Let z (0, ) be ot of the form / with N. Costruct a cotiuous fuctio f(x) defied o [0, ] such that 0 f(x) for all x [0, ], ad f(0) = f(), but there are o 0 x < y such that f(x) = f(y) ad y = x + z. Solutio: (a) Cosider the sequece of umbers ( ) ( k k α k = f f ), for k {, 2,, }. Notice that if α k = 0 for some k, the we are doe: x = k ad x 2 = k + So, let s assume all the α k s are ozero. satisfy the coditio. If α k < 0 for all k, the we have the chai of iequalities ( ) ( ) 2 f(0) > f > f > > f(), which is ot possible sice f(0) = f(). If α k > 0 for all k, the we have the chai of iequalities ( ) ( ) 2 f(0) < f < f < < f(), which is ot possible sice f(0) = f(). 5
It follows that ot all the α k ca have the same sig. Choose j such that α j ad α j+ have differet sigs. Cosider the fuctio g(x) = f(x) f ( x ) [ ; it is a cotiuous fuctio o ( ) ( ), ]. The by costructio, g j ad g j+ have differet sigs. It follows by the itermediate value theorem [ ] that there must exist y j, j+ such that g(y) = 0. This meas that f(y) = f ( y ). The, x = y ad x 2 = y satisfy the coditios. (b) The idea i the previous problem was to look at the sequece f ( k ). The fact that we get f() at some poit esured that this sequece coud ot be strictly icreasig. The, the fact aloe that this sequece is ot strictly icreasig allowed us to coclude. So, if we wat a couterexample, we eed the sequece f(kz) to be strictly icreasig. Similarly, we eed the sequece f( kz) to also be stricly icreasig. This suggests the followig example: defie partially a fuctio f by { j if z = jz for some 0 j f(x) = z j if z = jz for some 0 j z. This way, we have f(0) = 0 = f(). We kow that we did t assiged differet values to the same poit: if j z = j z for some j ad j, we would have z = j+j, but we assumed that z does ot have that form. Now defie f to be piecewise liear, by joiig together two successive poits already i the graph of f. Oe ca write a explicit formula, ad check that the coditio is satisfied. Or a picture would also do it. Of course this does t take values i [0, ]; it takes values i [ z, z ]. But addig z to the fuctio, ad the dividig the result by 2 z, makes it takes values betwee 0 ad. Problem 8: Let a fuctio f(x) be defied as follows: f(x) = 0 if x / Q, ad f(x) = /q if x = p/q with relatively prime itegers p ad q with q > 0, ad f(0) =. Show that f is cotiuous at irratioal x ad discotiuous at ratioal x. Solutio: We prove first that this fuctio is discotiuous at ratioal umbers. Let p q Q. The x = p q 2 is a sequece i R \ Q (see HW, Problem 2.(ii)) covergig to x 0 = p q. Sice x R \ Q, we have f(x ) = 0. So, lim f(x ) = 0 q = f(x 0), provig that f is ot cotiuous at x 0 = p q. Now, let r R \ Q be a irratioal umber; we wat to prove that f is cotiuous at that poit. Let ε > 0; we wat to fid δ > 0 (this δ depeds i geeral o both r ad ε) such that for every x with x r < δ, we have f(x) f(r) < ε. Choose N such that N < ε. For every k N, let δ k = mi{ r a k a Z}. Cosider the iterval [r, r + ]. It is clear that there are oly fiitely may multiples of k i this iterval. Sice r is irratioal, is it therefore clear that δ k > 0. Now, let δ = mi{δ, δ 2,, δ N, δ N } > 0 (because it is the miimum of a fiite collectio of positive umbers). By costructio, we are sure that o umber i the iterval (r δ, r + δ) is a multiple of k for k N. Now, let x such that x r < δ. If x is irratioal, we have f(x) f(r) = 0 0 = 0 < ε. If x is ratioal, sice it is i the iterval (r δ, r + δ), its deomiator has to be bigger tha N. Therefore, f(x) < N, ad we have f(x) f(r) = f(x) < N < ε. This proves that f is cotiuous. 6