Question

EXPLAIN STEP BY STEP

In Exercises 13 through 18 determine if the set of vectors S forms a subspace of the given vector space. Give reasons why S either is or is not a subspace.

Answer 1

= So UES. Now 14 s={(2,3...XnER" | x= x Let = (1, 1, 1, ...) Since 1 = 1 2u = 2(1,1,.... 1) (2,2,.... 2) Since 27 24 so 20&s This shows that s is not a subspace of IR? s={(2,5X2,... L) EIR" | 2+287..t nx=0} Let u= (14, 2, .. . Xn) ES and v= · (Y,I,... Yn) ES. Then x + 2x2+3x₂+... tnxn=0 y+2y, +34 + .. ....tryn=0 Now Utv = (x, dz....X) + Ly,,Y₂,-. Yn) =(x+y,, zut Y2, ... bat yn) Since wity + 2(x2+42) + 3 (x3 + Y3) + ... + (xntyn) = Qu +22+3x3+ +nan) +(4+244 +nyn) = 0+0 = 6 So utvES Let DER. Then Au=x (4,22,- = (4x4, dag, ... ,xn) Since da + 2ax₂ + 3 anzt... thxn =A(24+2 x₂ + 3x₃ + +nan) = 2.0 and - 0 so du Es This shows that s is a subspace of IR.

© S = {(24.2.2. ... ,xn) € IR" | *+xt tan=2 xn=2} Let u= (1,1,0,0,.. Since 1+1 to tot ...0 = 2,50 UES so ves utv Let v = (1,0,1,0,0,0) Since 1+0+1+0,70+. ..+0=2, (1,1,0,0....0) + (1,0,1,0,0,....) (2,1,1,0,0,...o Since 2+1+1+0+0+... +0=4 &S This shows that I is not a subspace of IR". 3 so utv