# Proof that there exists a polynomial of at most degree $2014$ so that for every $k \in \{1,2,…,2014\}$ $f_k(k)=0$ with f not the zero function.

I was studying for an exam and then I stumbled upon this problem which stumped me. I'd love if anyone would be able to proof this.

Let $\mathbb{R}[X]$ be the vector space of polynomials over $\mathbb{R}$ For every polynomial $f \in \mathbb{R}[X]$ we define $f_1 = f$ ,$f_2 = x \cdot f'$, $f_3 = x \cdot f_2'$...

Proof that there exists a polynomial of at most degree 2014 so that for every $k \in \{1,2,...,2014\}$ $f_k(k)=0$ with f not the zero function.

I tried constructing such a polynomial and tried things with sums of products of $(x-k)$ but it got really messy real quick. How does one easily proof this using linear algebra?

• why is this tagged linear algebra ? – KonKan Mar 6 '16 at 2:33
• @KonstantinosKanak0glou well I suppose one should be able to prove this using linear algebra, considering it was a problem in an old linear algebra exam I was making. – pokemonfan Mar 6 '16 at 2:36

A polynomial of degree at most 2014 is determined by 2015 coefficients. Each of the conditions $f_k(k) = 0$ is a linear condition on the coefficients. You are essentially asked to show that the null space of a $2014 \times 2015$ matrix is nontrivial.