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?

  • $\begingroup$ For $P(x) = x^n$ you get the Vandermonde matrix. $\endgroup$ Nov 30, 2010 at 17:25
  • $\begingroup$ @Yuval: You're right. For the general case directly calculating the determinant doesn't seem very nice, though. $\endgroup$
    – J. J.
    Nov 30, 2010 at 17:34

4 Answers 4


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).

  • 1
    $\begingroup$ I think you meant $P(x) = x^2$ in the example, so it took me a while to understand your argument. I think I got it, though. Just to check: The original linear dependency of the translates would appear as linear dependency in the rows of that matrix, but then by looking at the columns (and since row rank is col rank) you conclude that they must actually be independent? $\endgroup$
    – J. J.
    Nov 30, 2010 at 18:09
  • $\begingroup$ Yep, that's the idea. $\endgroup$ Nov 30, 2010 at 18:21
  • $\begingroup$ The tricky part is actually to ensure that the columns cannot have a non-trivial linear combination that equals zero. I tried to consider the polynomial $P(x+y)$ and evaluate it for $x=0,1,\dots,n$ to get the matrix, but then the argument fails. $\endgroup$ Feb 15, 2011 at 21:03

Here a fancy proof that uses operators and differentiation.

Let $T_a$ be the translation operator which maps polynomials to their translates

$$ T_a p(x) = p(x + a) .$$

Restricting attention to polynomials of degree $n$ or lower, this is a map from an $n+1$-dimensional vector space to itself. The question to be resolved is whether the $n+1$ translation operators $T_{a_0}, T_{a_1}\dots, T_{a_n}$ are linearly independent (when applied to polynomials of degree $n$).

It is well-known that the translation operator can be written as the exponentiation of the differentiation operator $Dp(x) := p'(x)$ as follows

$$ T_a = e^{aD} = 1 + \frac1{1!} aD + \frac1{2!} a^2D^2 + \dots + \frac1{n!} a^n D^n .$$

This is Taylor's theorem. Note that since $D$ is nilpotent on the space of polynomials of degree $n$, $D^{n+1} = 0$, the series is actually a finite sum.

Now, the point is that for any polynomial $p(x)$ of degree $n$, we know that its derivatives $p(x), Dp(x), D^2p(x), \dots, D^n p(x)$ are linearly independent because the degree always decreases by 1. Hence, they form a basis of our vector space of polynomials. The expansion of the $T_a$ in terms of this new basis is given by the expression above.

To show that the translates $T_{a_k} p(x)$ are linearly independent, we only have to check that their matrix with respect to the new basis is nonsingular. It reads

$$\begin{pmatrix} 1 & 1 & \dots & 1 \\ a_0 & a_1 & \dots & a_n \\ \frac1{2!} a_0^2 & \frac1{2!} a_1^2 & \dots & \frac1{2!} a_n^2 \\ \vdots \\ \frac1{n!} a_0^n & \frac1{n!} a_1^n & \dots & \frac1{n!} a_n^n \\ \end{pmatrix}$$

But its determinant is clearly a non-zero multiple of the Vandermonde determinant, which is non-zero if and only if the values $a_0,a_1,\dots,a_n$ are pairwise distinct.

(Another way to see that this matrix is non-singular is to note that the Vandermonde determinant is related to the interpolation problem, whose solution is unique.)


If you write $P(x) = \sum_{j=0}^n p_j x^j$ and set $ \sum_{i=0}^n c_i P(x+a_i) = 0$, the full expression is written $\sum_{i=0}^n \sum_{k=0}^n \sum_{j=0}^k c_i p_k {k \choose j} a_i^{k-j} x^j$. If we find the coefficient of each power $x_j$ as a matrix times the vector $(c_0, ..., c_n)$, and then setting the system equal to zero, we can deduce the determinant of the matrix is zero. The matrix is $A_{ij} = \sum_{k=j}^n p_k {k \choose j} a_i^{k-j}$.

Setting the determinant equal to zero gives a polynomial in all $a_i$. I don't know if this polynomial is simpler than the matrix hints at, but it is certainly possible to find coefficients $a_i$ such that the polynomials are linearly independent, though for "most" values it won't be the case, as the solution set must be of at least one dimension less than $C^{n+1}$ (seen by fixing all $a_1, ..., a_n$ arbitrarily and thereby determining a finite set of possible $a_0$).

  • $\begingroup$ Setting the determinant to zero will give the solutions where the polynomials are linearly dependent. What I think is that in each of these solutions there must exist $i \neq j$ such that $a_i = a_j$. I think you got your "most" the wrong way around - for "most" values it will be linearly independent. $\endgroup$
    – J. J.
    Nov 30, 2010 at 18:25
  • $\begingroup$ Oh yeah I forgot I was looking at the opposite case, thanks. $\endgroup$
    – chroma
    Nov 30, 2010 at 19:56

Here a fancy proof by induction that uses differentiation.

Case $n=0$.

A polynomial of degree $0$ has the form $P(x)=c_0$ with $c_0\neq 0$ which is clearly linearly independent.

Case $n=k+1$.

Assume that there is a linear dependence

$$\lambda_0 P(x+a_0) + \lambda_1 P(x+a_1) + \dots + \lambda_n P(x+a_n) = 0.$$

Since this relation holds for all $x$, we can differentiate this relation. After differentiating $n$ times, we obtain a relation for the $n$-th derivative

$$\lambda_0 P^{(n)}(x+a_0) + \lambda_1 P^{(n)}(x+a_1) + \dots + \lambda_n P^{(n)}(x+a_n) = 0.$$

But since $P$ has degree $n$, we know that $P^{(n)}(x+a_i)$ is equal to a constant $c\neq 0$, which implies

$$ \lambda_0 + \lambda_1 + \dots + \lambda_n = 0 .$$

By expressing $\lambda_0$ in terms of the other coefficients, we obtain a relation

$$ \lambda_1 (P(x+a_1) - P(x+a_0)) + \dots + \lambda_n (P(x+a_n) - P(x+a_0)) = 0$$

of polynomials of degree $n-1$. By induction, we conclude that all the $\lambda_i$ must vanish.

  • 2
    $\begingroup$ In the last step the polynomials are not necessarily of degree $n-1$. (Consider for example the situation where $a_1, \dots, a_n$ are all roots of $P$.) Moreover, what do you assume from the polynomials? The polynomials in the last step are not of the same form. (They are not necessarily translates of some other polynomial, are they?) $\endgroup$
    – J. J.
    Nov 30, 2010 at 19:39
  • $\begingroup$ Oops, sorry for my mistakes, and sorry for not noticing your comment until now. (!) I don't think the argument can be salvaged, but I have a new one in mind which is just as fancy, but hopefully works. $\endgroup$ Feb 16, 2011 at 17:31

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