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For each of the following sets, either verify (as in Example1) that it is a vector space, or show which requirements are not satisfied. If it is a vector space, find a basis and the dimension of the space.

Polynomials of degree7but with all odd powers missing.

Short Answer

Expert verified

The basis of the vector space: 1,x2,x4,x6.

Dimension of the vector space: 4

Step by step solution

01

Given information.

Polynomial of degree 7 with aodd=0.

02

Condition for vector space.

LetVbe a non-empty set and addition is an internal binary operation which maps VVV which assigns to each order pair (x,y) VVto a unique element of Vdenoted by x+yand multiplication is an external binary operation over the field Fthat maps FVV which assigns to each order pair (,x)FVto a unique element in V denoted by .x=伪虫 , then is called the vector space if it satisfies the following conditions:

(V,+)is an Abelian group

(x+y)=伪虫+伪测鈭赌x,yVandF(+)x=伪虫+尾虫鈭赌xVand,F(尾虫)=(伪尾)x=(伪虫)鈭赌,FandxV1.x=x鈭赌xV

03

Verify that the following set is vector space or not. If it is, then find dimension and basis of the space.

As Polynomial of degree 7 is given so this polynomial can be written as,

Px=a0+a1x+a2x2+a3x3+a4x4+a5x5+a6x6+a7x7

If aodd=0, then

Px=a0+a2x2+a4x4+a6x6

From the example 1,

Add two polynomials of this form will give another polynomial of the same form.

Addition of algebraic expression is commutative and associative.

The "zero vector鈥 is the polynomial with all coefficient ai=0 and adding it to any other polynomial just gives that other polynomial. Also the additive inverse of a function f(x)is-f(x)and -f(x)+f(x)=0as required for a vector space.

Hence all the properties of vector space are satisfied so, this set is a vector space.

04

Find dimension and basis of the space.

Now the functions 1,x2,x4,x6span the vector space i.e. any element of the vector space can be written as a linear combination of them. This show that they are linearly independent.

Therefore, the functions 1,x2,x4,x6 form the basis for the vector space, and therefore the vector space has dimension 4.

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