I've been wondering about the following question:
Suppose that $P$ is a polynomial of degree $n$ with complex coefficients. Assume that $a_0, a_1, \dots, a_n \in \mathbb{C}$ are distinct. Are the polynomials $$P(x + a_0), P(x + a_1), \dots, P(x + a_n)$$ linearly independent?
[Math] Are translations of a polynomial linearly independent
linear algebrapolynomials
Best Answer
Consider the matrix of coefficients, where each row corresponds to an evaluation at a point, and each column to a power of $x$. Any linear dependency among the columns amounts to some degree~$n$ polynomial in the point, which is supposed to have $n+1$ roots. Since this can't happen, the polynomials are independent.
Example (almost Vandermonde) - $P(x) = x^2$: $$\begin{pmatrix} 1 & 2a & a^2 \\ 1 & 2b & b^2 \\ 1 & 2c & c^2 \end{pmatrix}$$ A linear dependency of the columns is a quadratic polynomial having three different roots $a,b,c$.
EDIT: Just to clarify what happens in the general case. Here's the matrix corresponding to $P(x) = x^2 + \alpha x + \beta$: $$\begin{pmatrix} 1 & 2a + \alpha & a^2 + \alpha a + \beta \\ 1 & 2b + \alpha & b^2 + \alpha b + \beta \\ 1 & 2c + \alpha & c^2 + \alpha c + \beta \end{pmatrix}$$ Since the degree of the polynomial in column $t$ is $t$ (if we start counting from zero), it's easy to see that the any non-zero combination of columns results in a non-zero polynomial (consider the rightmost non-zero column).