Question

3. Find the dimension and give a basis for the vector space V {p(x) e P2| p(1) = 0}.

Answer 1

P = vector space of all polynomial of degree sax tbm te: a, b,CE R} V- { *(W) EP : 0 (1)=0}

Let Phan + bate EV Then PO = 0 -> a+b+c=0 Now P(x) = ant buto = anton-a-b = b (2-1) ta(x-1)

So P(M) E span{ x-1, x-1} and therefore v € Spang 2-1, =1} Again lef qm) . Again la Espans na ho ho CURSO for So q (W) = a, (1-1) +9, (2-1) Some an az E R. Now q (K) E PP2 and so que and therefore spanşx-1, x-12Ev. :: Va Spanfa-1, k-1}

Now we will show that {al, n-1} is linearly independent. Let t, pr-1) + 22 (2-1) = 0 This is true for all n. pulling nzo, we get -, -12 26-0 Pulling a=-1 we get -22, to=0 =>,=0 from ☺ , zo

Since (I) holds implies that do = 0, y =0, {x-1, x²-1} is a basis of V, and since the set contains 2 vectors, dim V=2.