Question

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 degree$\mathbf{\le }\mathbf{7}$but with all odd powers missing.

Short answer

The basis of the vector space: $\left\{1,{\mathrm{x}}^2,{\mathrm{x}}^4,{\mathrm{x}}^6\right\}$.

Dimension of the vector space: 4

Step-by-step solution

Given information.

Polynomial of degree $\le 7$ with ${\mathrm{a}}_{\mathrm{odd}}=0$.

Condition for vector space.

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

(V,+)is an Abelian group

$$\begin{gathered} \mathbf{\alpha }\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{)}\mathbf{=}\mathbf{\alpha x}\mathbf{+}\mathbf{\alpha y}\mathbf{\forall }\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{\in }\mathbf{V}\mathbf{}\mathbf{and}\mathbf{}\mathbf{\alpha }\mathbf{\in }\mathbf{F} \\ \mathbf{(}\mathbf{\alpha }\mathbf{+}\mathbf{\beta }\mathbf{)}\mathbf{x}\mathbf{=}\mathbf{\alpha x}\mathbf{+}\mathbf{\beta x}\mathbf{\forall }\mathbf{x}\mathbf{\in }\mathbf{V}\mathbf{}\mathbf{and}\mathbf{}\mathbf{\alpha }\mathbf{,}\mathbf{\beta }\mathbf{\in }\mathbf{F} \\ \mathbf{\alpha }\mathbf{\left(}\mathbf{\beta x}\mathbf{\right)}\mathbf{=}\mathbf{\left(}\mathbf{\alpha \beta }\mathbf{\right)}\mathbf{x}\mathbf{=}\mathbf{\beta }\mathbf{\left(}\mathbf{\alpha x}\mathbf{\right)}\mathbf{\forall }\mathbf{\alpha }\mathbf{,}\mathbf{\beta }\mathbf{\in }\mathbf{F}\mathbf{}\mathbf{and}\mathbf{}\mathbf{x}\mathbf{\in }\mathbf{V} \\ \mathbf{1}\mathbf{.}\mathbf{x}\mathbf{=}\mathbf{x}\mathbf{\forall }\mathbf{x}\mathbf{\in }\mathbf{V} \end{gathered}$$

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 $\le 7$ is given so this polynomial can be written as,

$\mathrm{P}\left(\mathrm{x}\right)={\mathrm{a}}_0+{\mathrm{a}}_1\mathrm{x}+{\mathrm{a}}_2{\mathrm{x}}^2+{\mathrm{a}}_3{\mathrm{x}}^3+{\mathrm{a}}_4{\mathrm{x}}^4+{\mathrm{a}}_5{\mathrm{x}}^5+{\mathrm{a}}_6{\mathrm{x}}^6+{\mathrm{a}}_7{\mathrm{x}}^7$

If ${\mathrm{a}}_{\mathrm{odd}}=0$, then

$\mathrm{P}\left(\mathrm{x}\right)={\mathrm{a}}_0+{\mathrm{a}}_2{\mathrm{x}}^2+{\mathrm{a}}_4{\mathrm{x}}^4+{\mathrm{a}}_6{\mathrm{x}}^6$

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” is the polynomial with all coefficient ${\mathrm{a}}_{\mathrm{i}}=0$ and adding it to any other polynomial just gives that other polynomial. Also the additive inverse of a function $\mathrm{f}\left(\mathrm{x}\right)\mathrm{is}-\mathrm{f}\left(\mathrm{x}\right)$and $-\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{x}\right)=0$as required for a vector space.

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

Find dimension and basis of the space.

Now the functions $\left\{1,{\mathrm{x}}^2,{\mathrm{x}}^4,{\mathrm{x}}^6\right\}$span 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 $\left\{1,{\mathrm{x}}^2,{\mathrm{x}}^4,{\mathrm{x}}^6\right\}$ form the basis for the vector space, and therefore the vector space has dimension 4.