Let W be the set of all vectors p(t) in

stop2dance3l

stop2dance3l

Answered question

2021-12-17

Let W be the set of all vectors p(t) in P such that p(0)=0. Show that W is a subspace of P and find a basis for W.

Answer & Explanation

Thomas White

Thomas White

Beginner2021-12-18Added 40 answers

Solution:
W be the set of all vectors p(t) in P2 such that p(0)=0
P2(t)={a0+a1t+a2t2a0,a1,a2∈R}
now, W={p(t)=a0+a1t+a2t2p(0)=0}
p(0)=0
a0+a1(0)+a2(0)=0
a0=0
W={a1t+a2t2a1,a2∈R}
check subspace condition,
1) let x=a1t+a2t2 and y=b1t+b2t2;x,y∈W
x+y=(a1t+a2t2)+(b1t+b2t2)
=(a1+b1)t+(a2+b2)t2
x+y∈W
2) x=a1t+a2t2∈W and a∈F be a scalar.
ax=a(a1t+a2t2)
=(aa1)t+(aa2)t2
ax∈W
3) put a1=a2=0
we get, p(t)=0
W has zero vector
Therefore W is a subspace of P2
now, W={a1t+a2t2a1,a2∈R}
=span{
Natalie Yamamoto

Natalie Yamamoto

Beginner2021-12-19Added 22 answers

A polynomial p(x)=ax2+bx+c satisfies p(1)=0 if and only if a+b+c=0
Hence every p∈W is of the form
p(x)=ax2+bx+(−a−b)=a(x2−1)+b(x−1) (1)
Can you use equation (1) to find a basis for W and thus compute dimW?
Sidenote: There is a more fun way to do this problem. The map T:P2⇒R given by T(p)=p(1) is linear and kerT=W. Since T(1)=1 it is clear that T has rank 1. The Rank-Nullity theorem then gives dimW.
nick1337

nick1337

Expert2021-12-28Added 777 answers

a.Since (1)=0, x-1 is a factor of P. So any P in W is of the form: . So if P and S are two elements in W, then, (P+S)(1)=0+0=0. So P+S∈W, and rP(1)=r⋅0=0. So rP∈W for any real number r. So W is a subspace of P2
b. To get a dimension of W we find a spanning set T of W and independent as well. Consider T={x−1,(x−1)2}. T is independent. We need to show T spans W. Let P be any element of W. So P(x)=(x−1)⋅q(x). If q(x) is a scalar, say q(x)=m,then P=m(x−1)+0(x−1)2, and if q(x) is not a scalar, then q(x)=ax+b for some number a,b. So P(x)=(x−1)(ax+b)=(b−a)(x−1)+a(x−1)2. So T spans W. Thus: T is a basis with 2 elements. So dim(W)=2

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