Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $L$ be a Vector Space of Polynomials of $x$ of degree $\leq n-1$ with coefficients in the field $\mathbb{K}$.

Define $$g_i(x) :=\prod _ {{j=1},{j\neq i}}^n \frac{x-a_j}{a_i-a_j}$$ Show that the polynomials $g_1(x), g_2(x),...,g_n(x)$ form a basis of L. Furthermore, show that coordinates of polynomial $f$ in this basis are $\{f(a_1),f(a_2),...,f(a_n)\}.$

To show that the polynomials are the bases, I need to show that they span $L$ and that they are linearly independent. I thought showing that any element in the set $\{1,x,x^2,...,x^{n-1}\}$ belongs to the span of $\{g_1(x), g_2(x),...,g_n(x)\}$ would be enough to show the $g_1(x), g_2(x),...,g_n(x)$ spans $L.$ But I don't know how to do this! Also, linear independence seems to be tougher!

share|cite|improve this question
What is $g_1$? You could use the fact that the size (cardinality) of any basis of $L$ is $n$. Then show that any function $x \mapsto x^k$ can be written in terms of $g_1,..,g_k$. Induction would work nicely here... – copper.hat Sep 22 '13 at 17:06
The $a_j$ are fixed elements from $K$? Right? – leo Oct 22 '13 at 2:20
up vote 3 down vote accepted

Here is very easy method to show that they are linearly independent.

Suppose that $$b_1g_1(x)+b_2g_2(x)+...+b_ng_n(x)=0$$ To show linear independence, it suffices to show that $$b_1=b_2=...=b_n=0$$ Evaluate $$b_1g_1(x)+b_2g_2(x)+...+b_ng_n(x)=0$$ at $$x=a_i\;\;\;\;\forall i\le n$$

You can notice that $$g_i(a_j)=\begin{cases}0&i=j \\ 1 & i\neq j\end{cases}$$

It follows that $$b_1=b_2=b_3=...=b_n=0$$

To see that the coordinates are given as such, consider a general polynomial as above $$f(x)=c_1g_1(x)+c_2g_2(x)+...+c_ng_n(x)=0$$

Follow same thing as above, start substituting $x=a_i$, you will see that $$c_1=f(a_1)$$ and so on.

Best of luck. I hope I could help.

share|cite|improve this answer
Why is it enough to show that $a_1g_1(x)+a_2g_2(x)+\ldots+a_ng_n(x)=0$ implies $a_1=\cdots=a_n=0$? I mean when proving linear independence we have to prove that for arbitrary scalars $c_1,\ldots,c_n\in K$, $$c_1g_1(x) + \cdots + c_ng_n(x)=0$$ implies $c_1=\cdots=c_n=0$. – leo Oct 22 '13 at 2:39
I think @leo's objection is that you (inadvertently?) reused the $a_i$ to denote arbitrary scalars with $\sum a_ig_i=0$. – Pedro Tamaroff Oct 22 '13 at 3:00

Choose arbitrary $f\in L$. Let be $$\tilde{f}(x) = \sum_{i = 1}^{n}f(a_i)g_i(x)\text{.} $$

For every $x\in \{a_1,\dots, a_n\}$ we have $f(x) = \tilde{f}(x)$, so the polynomial $p= f - \tilde{f}$ has $n$ zeros and $\deg p \leq n-1$, so $p(x) = 0$ for every $x\in \mathbb{R}$. So $g_i$ span $L$. We know that $\dim L = n$, so they must be linearly independendent.

share|cite|improve this answer

You missed the hypothesis that the $a_i$ are pairwise distinct. Alternatively, consider the functionals $\eta_i$ with $P\mapsto P(a_i)$ in $V=(\Bbb R_{n-1}[X])^\ast$, and suppose $$\sum_{i=1}^n\lambda_i\eta_i=0$$

Evaluating at $1,X,X^2,\ldots,X^{n-1}$ we get $n$ equations $k=0,1,2,\ldots,n$.


Or $$\begin{pmatrix}1&1&\cdots&1\\a_1&a_2& \cdots&a_n\\\vdots&\vdots &\ddots&\vdots\\a_1^{n-1}&a_2^{n-1}&\cdots&a_n^{n-1}\end{pmatrix}\begin{pmatrix}\lambda_1\\ \lambda_2\\\vdots\\ \lambda_n\end{pmatrix}=\begin{pmatrix}0\\ 0\\\vdots\\ 0\end{pmatrix} $$

But since the $a_i$ are pairwise distinct, the Vandermonde matrix is invertible, which means ${\bf \lambda} =\bf 0$ as desired.

The claim follows from the fact your polynomials are precisely the predual basis, call it $B$, of the evaluations above which form a basis $B'$ for $V$. This observation also means that $$(f)_B=(\eta_1f,\ldots,\eta_n f)=(f(a_1),\ldots,f(a_n))$$

This is a general property: if $B^*=\{\varphi_1,\ldots,\varphi_n\}$ is the dual basis of $B=\{v_1,\ldots,v_n\}$, and if $v\in V$, $\varphi\in V^\ast$, then $$(v)_B=(\varphi_1(v),\ldots,\varphi_n(v))$$ $$(\varphi)_{B^\ast}=(\varphi(v_1),\ldots,\varphi(v_n))$$ and the proof is just like that exhibited in the accepted answer: evaluate, and use $\varphi_i(v_j)=\delta_{ij}$.

share|cite|improve this answer

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.