Question

Let V be a vector space and W be a subset of V. By justifying your answer, determine whether W is a subspace of V. (a) V = M2,2 and EV : ad = bc (b) V = P3 and W = {a + bx + cx? + dx€ V: a+c= b+d}. +

Answer 1

a rector Then i subiet of vis space said w of v is to be dubspace of v if Zero nector is in in W cio Cui) W is closed under addition that is it و کار ما EW then Wito, EW (ii) W is W is closed under scalar multiplication. that is a is scalar and wew it then XWEW

and (a): V = m २,२ W= d not w is Here Because Leevadare? a subspace of [:] [: ;) [:]+[• :) [. ( Here ad + bc) As EW But ;] &W Here (0) (0) (0) -a subspace of v wi NOT

16) V=P3 and W= { a+bx+ cx?+dx? EV : a+c=67413 vector in w 2+ox? 0+0:17 ton AS Zero Oto - Opo. for this i in W Hence zero vector and Now Let w = atbrt cx²+ din? Wa= a + bg x + x² + da x² EW W,,W2 3 het 3 3 Witwa Hence a+bet ar?+ dix? + azt baht Ca ut dan ai tag + (6; tb;)n + (6+G) x + (derda] X here low,,We EW at az + G + C a ait G+ atą bit dit barda - ata=bito, : b, tbat detda (and az + G = bgt de Witwg tw

of ņ GW W,, Wa WitW GW - W. is closed under adelition. Now for لح 3. for w= at but ex²+ da a ý a scalar and Here WEW => xw = a (a+bx+ cx?t dr3) = da + ({ b) x + (*C )x++/dd) x3 Now da + C = a (a + c) alb+d) L*WEW sa+c=6+d Ia btad بال xwEW XWE W and a is scalar Hence if wew =) Hence w is closed under scalal multiplication. W is a subspace of v.