# Show that the set is a vector space in the field of real numbers?

## $V$ $=$ {p∈ p_3(x)|p(0)=0,p´(0)=0}

Jun 24, 2018

See below

#### Explanation:

Let's check all the requirements:

• The set must be closed with respect to the sum. This is true because, given two polynomials of degree $3$, we have

${a}_{0} + {a}_{1} x + {a}_{2} {x}^{2} + {a}_{3} {x}^{3} + {b}_{0} + {b}_{1} x + {b}_{2} {x}^{2} + {b}_{3} {x}^{3}$

$= \left({a}_{0} + {b}_{0}\right) + \left({a}_{1} + {b}_{1}\right) x + \left({a}_{2} + {b}_{2}\right) {x}^{2} + \left({a}_{3} + {b}_{3}\right) {x}^{3}$

which is still a polynomial of degree $3$. Moreover, if ${p}_{0} = 0$ and ${p}_{1} ' \left(0\right) = 0$ and the same goes for ${p}_{2}$, we have

$\left({p}_{1} + {p}_{2}\right) \left(0\right) = {p}_{1} \left(0\right) + {p}_{2} \left(0\right) = 0 + 0 = 0$

and

$\left({p}_{1} ' + {p}_{2} '\right) \left(0\right) = {p}_{1} ' \left(0\right) + {p}_{2} ' \left(0\right) = 0 + 0 = 0$

• Closure with respect to scalar multiplication: if $p \left(0\right) = 0$ and $p ' \left(0\right) = 0$, for every $k \setminus \in \setminus m a t h \boldsymbol{R}$ we still have

$\left(k p\right) \left(0\right) = k \setminus \times p \left(0\right) = k \setminus \times 0 = 0$

and similarly

$\left(k p\right) ' \left(0\right) = k \setminus \times p ' \left(0\right) = k \setminus \times 0 = 0$

All of the other requirements (commutative sum, distributive property with respect to the scalars,..) derive from the fact that the sum between polynomials behave as numeric sum, since we just manipolate the like terms as above.