$\mathbb{V}$ is the multitude of all polinomials of degree less than or equal to 3. In this problem, I have to:
- prove that $\mathbb{V}$ is a linear space over $\mathbb{R}$.
- find the dimension of $\mathbb{V}$
- prove that the polinomials $1, x-91, \frac{(x-91)^2}{2!}, \frac{(x-91)^3}{3!}$ form a basis of $\mathbb{V}$
- prove that the multitude of polinomials $f \in \mathbb{V}$, for which $f(91)=0$ form a linear subspace of $\mathbb{V}$
$\mathbb{V}$ is the multitude of all polinomials of degree less than or equal to 3. Therefore, $\mathbb{V}=F^4[x] \subset F[x]$. We can present it as$$F^4[x]=\{f(x)=a_0x^0+a_1x^1+a_2x^2+a_3x^3\mid a_0, a_1, a_2, a_3\in F\}.$$The standart basis for this linear space is $e_1=1, e_2=x, e_3=x^2, e_4=x^3$. So far, I have proven the first and second points. For the third point, I expressed the four polinomials using the standart vectors $e_1, e_2, e_3, e_4$ and formed the matrix$$\begin{pmatrix} 1&0&0&0\\\ -91 & 1 & 0 & 0\\\\ \frac{{91}^2}{2}&-91&\frac{1}{2}&0\\\\ -\frac{{91}^3}{6}&\frac{{91}^2}{2}&\frac{91}{2}&\frac{1}{6} \end{pmatrix}$$ The determinant of this matrix is non-zero, therefore the four polinomials are linearly independent, therefore, they form a basis of $\mathbb{V}$. The fourth point however is problematic for me. I can't think of any ideas how to solve it. Thank you in advance!
If $f(91)=g(91)=0$, then$$(f+g)(91)=f(91)+g(91)=0.$$And, if $\alpha\in F$ and if $f(91)=0$,$$(\alpha f)(91)=\alpha f(91)=0.$$So, that set is indeed a subspace.