Some linear independence proof involving real numbers 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 $i$.
 A: 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 $\lambda = 0$.
A: 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.
A: 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).
