Answer:
Yes. It is a vector space over the field of rational numbers [tex]\mathbb{Q}[/tex]
Step-by-step explanation:
An element [tex]p[/tex] of the set [tex]H[/tex] has the form
[tex]p(x)=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}[/tex]
where [tex]a_{0},a_{1},a_{2},a_{3},a_{4}[/tex] are rational coefficients.
The operations of addition and scalar multiplication are defined as follows:
[tex]p(x)+q(x)=(a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+x_{4}x^4)+(b_{0}+b_{1}x+b_{2}x^{2}+b_{3}x^{3}+b_{4}x^{4})=(a_{0}+b_{0})+(a_{1}+b_{1})x+(a_{2}+b_{2})x^{2}+(a_{3}+b_{3})x^{3}+(a_{4}+b_{4})x^{4}[/tex]
[tex]\lambda p(x)=\lambda (a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4})=\lambda a_{0}+\lambda a_{1}x+\lambda a_{2}x^{2}+\lambda a_{3}x^{3}+\lambda a_{4}x^{4}[/tex]
The properties that [tex]H[/tex], together the operations of vector addition and scalar multiplication, must satisfy are:
This is not difficult with the definitions given. The most important part is to show that [tex]H[/tex] has a additive identity, which is the zero polynomial, that is closed under vector addition and scalar multiplication. This last properties comes from the fact that [tex]\mathbb{Q}[/tex] is a field, then it is closed under sum and multiplication.