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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.

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up vote 4 down vote accepted
  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 $P(X+n)\in\Bbb R_{n-1}[X]$.

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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.

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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.

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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.

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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! – Dan 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

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