Question

row reduction in uncountable dimension.

Part 2. (Row-reduction in countably-infinite dimension) Let V denote the vector space of polynomials (of all degrees). Recall that V is an infinite-dimensional vector space, but it has a countable basis. Consider Te Hom(V, V) defined as T(p())5p () 10p(x - 1) 2.1. Write T as an oo x oo matrix, in the standard basis 1,X, x2, 13,... of V 2.2. Write T as an oo x oo matrix, in the basis 1, + 1, (1), (a1)*, ...of V 2.3. Find the kernel of T. That is, find all polynomials that satisfy p'() = -2p(a - 1. (Hint: Even though V contains polynomials of arbitrarily high degree, each individual polynomial has a particular finite degree.) 2.4. Find all polynomials that satisfy the inhomogeneous system 5p (r)10p(x - 1) = 53. (Hint: Think of this as T(p) b. You have already solved T(p) 0.) +

Answer 1

Let denote the Vectoy polynomials, Space of (of degrees) but V finite dimens ional vectov space is has bas is. covntable Consider defined aS TEHom(V,V) T (рся)) : 5 px)1opix-) the a standaxd basis , х, х2 V. S d) 0 (1) - 5 (o) + 10 = 10 dx T(1) \<& A n-o 5 X ti0Cy (1-2 Y-2 +5-10) (- Cn 1-3 nco

wte T a n хс matix standard о.у basis оf then this columns matrix Ore Соefficients оf Сanm Let хс оf then Coeficient an.m aр.m: 0 - U As Ti10 JOх8- 5%x T) Х-2 U-A an, O m> 2 Amm 10, mt-Cn On.m 7 Сn, (m (алм). hen оо Х 0- О n22 amm 10, Qm,m 5m-) Gn,m (O an, m O m22, Mn iSman "Cn) U7w

2-2 Let (bnm w.Y.t basis сСх ос V Similar to 2.1. Гоefficient T(x ) TD с10. 5 d (t1 dя T (x1 u-е But СХ+1-13% 2 T( - 10. Всxt-19. (%н) (5x \-2 х + -Y с u-я С#1) . Вc,t-1) 10 nEo 5 8 10- (хн8 7-2

TU A om.m 10, bmm -5m ) m2 2. bnm lo. m (-) n- IO m V-w (- Со,som As n -i Caefficient 0f bn.m bn,m basis bom)x w.Y. V thea of n 2 bm,m0, bm1,- -S(m-1 n m bn,m 2 3 prx)2 plx-1) polynomial of Ate As desvee deg (pt) Let

then pu) ao (a Constant ) ten C learly -2p(x-)- 2ao 2aoO - px)-2plx-)., ihen & polyaomial ze ro Suppose nd!; > degresof As deg (px) -n pix) is of CoefRicient not eqval anx' Let px) an deg(px)n) ano 2 plx-) P'x) n-2 n- an-". n an a

paring Com coefRrients both in R. S and we gef Bot an 0 mption This the contra dicts deg p) that deg(p) prx)0 if pex)2p(x), then pox)0 Kernel T 2) Sp')Op-i) Sx3 T(p) Sx3 0(x- T3 S3x2) 2.1-, 1Ox3-S+ 30 x -10 TOX2 5(2x) Ox-5 Tx)5oCx-t)

Tx)T ( 10x3Sx io) 3 10x-101) 2. 2 - 2 SA + X gl. Tx) 국 (10x-5) Tx3)T2) 5x3S - x 3 25 SX T) 3 5 x°- H 3 x4- t then pix) 3+ s pru)op(x-i) 5x- 2 3 S Also One One Key(T) o T As the only S 2 H for Solution sp )to ptx) - Sx3 nloo