Let S be the subspace of P3 consisting of all polynomials px

Let S be the subspace of P_3 consisting of all polynomials p(x) such that p(0) = 0, and let T be the subspace of all polynomials q(x) such that q(1) = 0. Find bases for (a) S (b) T (c) S T

Solution

a)

as given S be the subspace pf P3 consisting pf all polynomials p(x) suxh that p(0) = 0

consider that S is all polynomials of the of the p(x) = ax² + bx

also a, b is the real numbers

it is like this becuase p(0) = a(0)² + b(0) = 0foe every a, b.

now consider that {x, x²} forms a basis for S.

so we need to show that two vectors x and x² are linearly independent and span S.

so we need to show that

c₁(x²) + c₂(x) = 0(x²) + 0(x)

so it has only one solution c₁ = c₂ = 0.

if we grouping the terms we find that : c₁ = c₂ = 0.  

hence the two vectors are linearly independent.

to show that the two vectors span S we need show that any element in S which represented by

p(x) = ax² + bx can be written as:

c₁(x²) + c₂(x) = ax² + bx.

c₁, c₂ are scalars.

if we grouping the terms we find that: c₁ = a and c₂ = b

hence we have : ax² + bx = ax² + bx

so two vectors span S.

as the vectors are linearly independent and span S means {x, x²} forms a basis for S.

b)

as given T be a subspace of all polynomials q(x) suxh that q(1) = 0

so T is all polynomials of the form

q(x) = a(x 1)(bx + c)

q(x) = abx² + acx abx ca

q(x) = ab(x²) + (ac ab)x ac

where a, b, c are real numbers.

it is like this becuase q(1) = a(1 1)(b + c) = 0 for every a, b, c.

For simplicity let s = ab and t = ac.

so T is gievn by q(x) = sx² + (t s)x t

consider that {(x 1),(x 1)²} forms a basis for T.

so we must show that the vectors x 1 and (x 1)² are linearly independent and span T.

for linearly independent we must show that:

c₁((x 1)²) + c₂(x 1) = 0(x 1)(0(x) + 0)

if we grouping the terms

we find: c₁ = 0 and c₂= 0

so the two vectors are linearly independent.

to show that the two vectors span T we must show that any element in T which is represented by

q(x) = sx² + (t s)x t can be written as:

c₁((x 1)²) + c₂(x 1) = sx² + (t s)x t.

where c₁ and c₂ are scalars.

so c₁(x² 2x + 1) + c₂(x 1) = sx² + (t s)x t.

c₁x² 2c₁x + c₁ +  c₂x c₂ =sx² + (t s)x t.

c₁x² +(c₂ - 2c₁₎x + c₁ c₂ =sx² + (t s)x t.

if we grouping the terms we find that:

c₁ = s

c₁ c₂ = -t

s - c₂ = -t

c₂ = t+s

c₁ = s and c₂ = s + t

so we have

sx² + (t s)x t = sx² + (t s)x t

which means the two vectors span T.

the two vectors are linearly independent and span T so {(x 1),(x 1)²} forms a basis for T.

c)

as S T is all polynomials of the form

c(x) = a(x1)(bx)

c(x) = abx²abx

a,b are the real numbers.

it is so that becuase c(0) = a(0 1)² (b(0)) = 0

and c(1) = a(1 1)² (b(1)) = 0 for every a, b.

for simplicity ab = t.

so S T is all polynomials of the form c(x) = tx²tx = tx(x1).

consider that {x(x 1)} forms a basis for S T.

so we must show that the vector x(x 1) is linearly independent and spans S T.

for linearly independent we show that :

d₁(x(x 1)) = 0(x(x 1)) only has only one solution d₁ = 0.

if we grouping the terms d₁= 0

so the vector is linearly independent.

Now to show that the vector spans S T

we must show that any element in S T which is represented by

c(x) = tx(x 1) can be written as

d₁(x(x 1)) = tx(x 1)

where d1 is a scalar.

if we grouping the terms we find that

d₁ = t

hence we have: tx(x 1) = tx(x 1)

it means the vector spans S T.

the vector is linearly independent and spans S T which means {x(x1)} forms a basis for S T