Question

Show that the polynomial ring $\mathit{R}\left[x\right]$(with the usual addition of polynomials and product of a constant and a polynomial) is a vector space over R.

Short answer

Answer:

$\mathit{R}\left[x\right]$is a vector space over R.

Step-by-step solution

Definition of space vector.

A space consisting of vector, together with the associative and commutative operations of addition of vectors and the associative and distributive operation of multiplication of vectors by scalars.

Showing that af(x)=∑i=0nαaixi ∈R[x]

Let $\mathit{R}\left[x\right]$be the ring of polynomial. Then the addition and the scaler multiplication be defined by $\mathit{a}\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{n}}{\mathit{a}}_{\mathbf{i}}{\mathit{x}}^{\mathbf{i}}\mathbf{,}\mathbf{}\mathit{g}\left(x\right)\mathbf{}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{n}}\mathbf{\in }\mathit{R}\mathbf{\left[}\mathbf{x}\mathbf{\right]}$and $\mathit{\alpha }\mathbf{\in }\mathit{R}$.

$\mathit{f}\left(x\right)\mathbf{+}\mathit{g}\left(x\right)\mathbf{=}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{n}}\left(a_i+b_i\right){\mathit{x}}^{\mathbf{i}}$

And

$\mathit{a}\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{n}}\mathit{\alpha }{\mathit{a}}_{\mathbf{i}}{\mathit{x}}^{\mathbf{i}}\mathbf{}\mathbf{\in }\mathit{R}\mathbf{\left[}\mathbf{x}\mathbf{\right]}$. Where $\mathit{m}\mathbf{=}\mathit{m}\mathit{a}\mathit{x}\left(n,p\right)$

Then apply the required condition to be a vector space over a field.

Step 3:Showing that α(fx+gx)=αf(x)+αg(x).

As $\mathit{R}\left[x\right]$is an abelian group with the respect to addition.

Let $\mathit{\alpha }\mathbf{(}\mathbf{f}\left(x\right)\mathbf{+}\mathbf{g}\left(x\right)\mathbf{)}\mathbf{=}\mathit{\alpha }\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\mathit{\alpha }\mathit{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{\alpha }\mathbf{,}\mathit{\beta }\mathbf{\in }\mathit{R}$.

Then,

$$\begin{gathered} \alpha \left(f\left(x\right)+g\left(x\right)\right)=\alpha \left(\sum _{i=0}^m\left(a_i+b_i\right)x^i\right) \\ =\sum _{i=0}^m\alpha a,x^i+\sum _{i=0}^m\alpha b_i{x}^i \\ =\alpha \sum _{i=0}^{m}a,x^i+\alpha \sum _{i=0}^m{b}_i{x}^i \end{gathered}$$

This gives,

Step 4:Showing that (α+β)f(x)=αf(x)+βf(x).

Third,

$$\begin{gathered} \mathbf{(}\mathit{\alpha }\mathbf{+}\mathit{\beta }\mathbf{)}\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{(}\mathit{\alpha }\mathbf{+}\mathit{\beta }\mathbf{)}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{n}}{\mathit{a}}_{\mathbf{i}}{\mathit{x}}^{\mathbf{i}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{n}}\left(\alpha +\beta \right){\mathit{a}}_{\mathbf{i}}{\mathit{x}}^{\mathbf{i}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{n}}\mathit{\alpha }{\mathit{a}}_{\mathbf{i}}{\mathit{x}}^{\mathbf{i}}\mathbf{+}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{n}}\mathit{\beta }{\mathit{a}}_{\mathbf{i}}{\mathit{x}}^{\mathbf{i}} \end{gathered}$$

This gives,

$\mathbf{(}\mathbf{\alpha }\mathbf{+}\mathbf{\beta }\mathbf{)}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathit{\alpha }\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\mathit{\beta }\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$.

Step 5:Showing that (αβ)f(x)=α[βfx].

$$\begin{gathered} \mathbf{\left(}\mathbf{\alpha \beta }\mathbf{\right)}\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\left(\alpha \beta \right)\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{n}}{\mathit{a}}_{\mathbf{i}}{\mathit{x}}^{\mathbf{i}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{n}}\left(\alpha \beta \right){\mathit{a}}_{\mathbf{i}}{\mathit{x}}^{\mathbf{i}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{\alpha }\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{n}}\mathit{\beta }{\mathit{a}}_{\mathbf{i}}{\mathit{x}}^{\mathbf{i}} \end{gathered}$$

Then gives

$\mathbf{\left(}\mathbf{\alpha \beta }\mathbf{\right)}\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathit{\alpha }\mathbf{\left[}\mathbf{\beta f}\left(x\right)\mathbf{\right]}$

Step 6:Showing that R[x] is a vector space in R.

$$\begin{gathered} I·f\left(x\right)=I·\sum _{i=0}^n{a}_i{x}^i \\ =\sum _{i=0}^{n}I·a_i{x}^i \\ =\sum _{i=0}^n{a}_i{x}^i \end{gathered}$$

This gives

$I·f\left(x\right)=f\left(x\right)$

All condition for $\mathit{R}\left[x\right]$to be a vector space is satisfied. Therefore polynomial right $\mathit{R}\mathbf{\left[}\mathbf{x}\mathbf{\right]}$is a vector space over$\mathit{R}$.