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If the equalities $$\lambda_0a_i⁰+...+\lambda_{n-1}a_i^{n-1}=0,\quad i=1,...,n$$ hold for fixed distinct real $a_i$, how can we conclude $\lambda_i=0$ for all $i$.

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    The polynomial $\sum_{j=0}^{n-1} \lambda_j x^j$ cannot have more than $n-1$ roots, because its degree is $n-1$. Hence it must be the zero polynomial. – Crostul Dec 21 '17 at 21:41

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The matrix $V$ with entries $[V]_{ij} = a_i^{j-1}$ is known as the Vandermonde matrix and has non zero determinant.

Hence if $\lambda=(\lambda_0,\cdots, \lambda_{n-1})^T$ and $V \lambda = 0$ we must have $\lambda = 0$.

copper.hat
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The relevant fact here is the following:

Theorem: if $$p(x) = \sum_{k = 0}^n \lambda_k x^k$$ is a non-trivial degree $n$ polynomial, then $p(x) = 0$ has at most $n$ solutions.

For a proof, see this other post on MSE. As your "polynomial" has $n$ roots but its degree is less than $n$, so it's the constant zero function.

egreg
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davidlowryduda
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This is much more cumbersome than @Crostul 's explanation, but maybe can help too

You have a linear system of the form $$\left(\begin{matrix} 1 &a_1 & a_1^2 &\cdots &a_1^{n-1}\\ 1 &a_2 & a_2^2 &\cdots &a_2^{n-1}\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 &a_n & a_n^2 &\cdots &a_n^{n-1}\\ \end{matrix} \right) \cdot \left(\begin{matrix} \lambda_0\\ \lambda_1\\ \vdots\\ \lambda_{n-1}\\ \end{matrix} \right) = \left(\begin{matrix} 0\\ 0\\ \vdots\\ 0\\ \end{matrix} \right) $$

So there is only the trivial solution ($\lambda_i=0$ $\forall i$) iff the squared matrix is invertible.

Now, that matrix (or its transpose, not sure but it doesn't matter) is usually called a Van der Monde matrix, let's say $V(a_1,a_2,\ldots,a_n)$. Using induction it can be proven that $$\det(V(a_1,a_2,\ldots,a_n))=\prod_{1\le i <j \le n} (\lambda_i - \lambda_j)$$ which is not zero iff all the $\lambda_k$ are different (and iff the matrix is invertible).

Alejandro Nasif Salum
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