Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $P \in \mathbb{R}_{n-1}[X]$ be a polynomial of degree $n-1 \geqslant 0$.

  1. Let $\mathbb{R}_{n-1}[X]$ be the vector space of polynomials with degree $\leqslant n-1$ over $\mathbb{R}$. Show that $(P(X),P(X+1),\ldots ,P(X+n-1))$ is a basis of $\mathbb{R}_{n-1}[X]$.

  2. Let $M_n = \begin{pmatrix} P(X) & P(X+1) & P(X+2) & \ldots & P(X+n) \\ P(X+1) & P(X+2) & P(X+3) & \ldots & P(X+n+1) \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ P(X+n) & P(X+n+1) & P(X+n+2) & \ldots & P(X+2n) \end{pmatrix}$.

Show that $\det{M_n} = 0$ for every $X \in \mathbb{R}$.

My thoughts on (1): $\mathbb{R}_{n-1}[X]$ is $n$-dimensional, because $(1,X, \ldots ,X^{n-1})$ is a basis of $\mathbb{R}_{n-1}[X]$. So it suffices to show that $(P(X),P(X+1),\ldots ,P(X+n-1))$ is a generating set/linearly independent. I tried proving it with induction and using the binomial theorem, but I am not getting anywhere.

My thoughts on (2): $\det{M_n} = 0$ implies that the columns are linearly dependent. (1) is probably useful here, but I don't even know how to start.

Any help is appreciated, thanks.

share|improve this question
add comment

4 Answers

  1. Induction is a good idea. I won't show the case $n=2$; assume that the result holds for $n-1\geqslant 1$ instead of $n$. The family $(P(X+k))_{k=0}^{n-1}$ generates the same subspace as $(P(X),(Q(X+k))_{k=0}^{n-2})$, where $Q(X)=P(X+1)-P(X)$. As $Q$ has degree $n-2$, $(Q(X+k))_{k=0}^{n-2})$ generates $\Bbb R_{n-2}[X]$. If $(\lambda_k)_{k=0}^{n-1}$ is such that $\lambda_0P(X)+\sum_{k=1}^{n-1}\lambda_kQ(X+k)=0$, then $\lambda_0=0$ by a degree argument, and by the induction hypothesis, $\lambda_k=0$ if $k\geqslant 1$.

  2. We can show that the columns are linearly dependent, as $P(X+n)\in\Bbb R_{n-1}[X]$.

share|improve this answer
add comment

For 1, your method to show by induction that it is a generating family is fine. Let $Q(X)=P(X+1)-P(X)$. Then $\deg Q=\deg P -1$ (if $n>1$). Indeed, if $P(X)=a_{n-1}X^{n-1}+a_{n-2}X^{n-2}+\ldots$, then $$\begin{align}&(a_{n-1}(X+1)^{n-1} &+ a_{n-1}(X+1)^{n-2}+\ldots) \\ -&(a_nX^{n-1}&+a_{n-1}X^{n-2}+\ldots) \\ =&(n-1)a_{n-1}X^{n-2}&+\ldots \end{align}$$ By induction hypothesis, $Q(X), \ldots, Q(X+n-2)$ generate $\mathbb R_{n-2}[X]$. Thus for any polnomial $f(X)=b_{n-1}X^{n-1}+\ldots\in\mathbb R_{n-1}[X]$, you can express $f(X)-\frac{b_{n-1}}{a_{n-1}}P(X)$ as linear combination of $Q(X), \ldots, Q(X+n-2)$, hence of $P(X), \ldots, P(X+n-1)$ and finally also $f$ as such linear combination.

share|improve this answer
I am struggling to understand the last step. $P(X),\ldots ,P(X+n-1)$ are linear combinations of $f(X),Q(X),\ldots ,Q(X+n-2)$, I get that. But I don't see why this implies that $f$ is a linear combination of $P(X), \ldots , P(X+n-1)$. Am I missing something here, a theorem or something rather obvious? –  user49797 Jan 3 '13 at 22:46
add comment

An easy approach will be to use the idea of Method of Finite Differences for a polynomial. For polynomial $F$ with degree $n$ and leading coefficient $\alpha$, the ith method of difference polynomial has degree $n-i$ and leading coefficient $n(n-1)\ldots (n-i+1) \alpha$

  1. For $i = 0$ to $n-1$, show that the ith method of difference polynomials are linearly independent, since each of them have a different degree. Use this to conclude that your polynomials are linearly independent.

    For clarity, these polynomials are $( P(X), P(X+1)-P(X), P(X+2)-2P(X+1)+P(X), \ldots)$. We may add multiples of the initial polynomials, and keep the set linearly independent still. Hence $(P(X), P(X+1), P(X+2), \ldots P(X+n-1))$ are linearly independent.

  2. The rows (and columns) satisfy the condition that their $n^{th}$ method of difference is equal to $0$. This gives you coefficients, which shows that the columns (rows) are not linearly independent, so the matrix has determinant 0.

    For clarity, the nth difference tells us that for any $j$, $$0 = {n\choose 0} P(X+j+n) - {n \choose 1} P(X+j+n-1) + {n\choose 2} P(X+j+n-2) - \ldots + (-1)^{n-1}{n\choose n-1}P(X+j+1) +(-1)^n P(X+j).$$ So take $n \choose i$ as the coefficients for the linear combination that is 0.

share|improve this answer
add comment

The above answers are great; I just thought I'd give a rather nice, alternative proof of the statement in question 1 which doesn't rely on induction.

I'll show just that any two of the basis elements are linearly independent. You can generalise it to show that the whole basis is a set of linearly independent elements.

Assume that $P(X)$ is non-constant. Then, $P(X+k)$ and $P(X+l)$ each have at most $n-2$ points with zero derivative (fundamental theorem of algebra). As they are translates of each other, there must exist some point $x_0$ such that $P'(x_0+k)=0$ but $|P'(x_0+l)|>0$.

Now, suppose there exist non-zero real numbers $n_1$ and $n_2$ such that $n_1P(X+k)+n_2P(X+l)=0$, then differentiating both sides, we get that $n_1P'(X+k)+n_2P'(X+l)=0$ for all $X$ in the real numbers. However, this includes $x_0$ and so $n_1P'(x_0+k)+n_2P'(x_0+l)=0$. Hence $n_2=0$ because $P'(x_0+k)=0$ and $P'(x_0+l)\neq 0$. But then either $n_1=0$ also, or $P(X+k)=0$ which can't happen because $P$ is non-constant. It follows that $P(X+k)$ and $P(X+l)$ are linearly independent.

share|improve this answer
I should comment that I'm now doubting how easily this argument generalises to larger sets of elements. In particular, I'm not entirely sure where the fact that we're taking a set of $n-1$ elements and where a set of $n$ elements stops being a basis. Take the above with a pinch of salt. If anyone can see how this generalises, please comment! –  Daniel Rust Jan 3 '13 at 22:27
It stops being a basis, since $P^{(n)}(X+k)=0$, so the equation will be satisfied. I think that you need to use the fact that $P^{(i)}$ is a polynomial of degree $i-1$, and jump through a bunch of induction hoops (which you were trying to avoid). Take a look at my solution, it doesn't involve induction, and uses the 'differentiation' idea. –  Calvin Lin Jan 3 '13 at 23:32
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.