I'm struggling with proving the Vandermonde matrix of dimension 4x4. I don't want to get into induction, if that is possible. I know there is a lot of material on the internet but I am looking for a calculation solution, and not an induction one. I have reached this expression:
$$ a_1^3(a_4-a_3)(a_3-a_2)(a_4-a_2)-a_2^3(a_4-a_3)(a_4-a_1)(a_3-a_1)+a_3^3(a_4-a_2)(a_4-a_1)(a_2-a_1)-a_4^3(a_3-a_2)(a_3-a_1)(a_2-a_1) $$ Obviously this is a very long expression to simplify.
Thanks in advance.
On
It's easier to prove the general formula, with some simple results on polynomials in several indeterminates. Consiider the Vandermonde determinant: $$\begin{vmatrix} 1&1&\dots&1\\ a_1&a_2&\dots&a_n\\ a_1^2&a_2^2&\dots&a_n^2\\ \vdots&&&\vdots\\ a_1^{n-1}&a_2^{n-1}&\dots&a_n^{n-1} \end{vmatrix}$$ The general formula for the determinant: $$\sum_{\sigma\in S_n}(-1)^{\varepsilon(\sigma)}c_{1\,\sigma(1)}c_{2\,\sigma(2)}\dotsm c_{n\,\sigma(n)}=\sum_{\sigma\in S_n}(-1)^{\varepsilon(\sigma)}1\cdot a_{\sigma(2)}\dotsm a_{\sigma(n)}^{n-1}$$ shows this determinant is homogeneous in the indeterminates $a_1, a_2, \dots,a_n$, with degree $$1+2+\dots+(n-1)=\frac{n(n-1)}2.$$ Now Vandermonde determinant is $0$ whenever $a_i=a_j$ for some $1\le i <j .\le n$. Hence it is divisible by $a_j-a_i$. As these polynomials are pairwise coprime, Gauß's lemma ensures it is divisible by their product $$(a_2-a_1)\dotsm(a_n-a_1)(a_3-a_2)\dotsm (a_n-a_2)\dotsm(a_n-a_{n-1}).$$ Observe this product has precisely degree $\dfrac{n(n-1)}2$. Hence it is a constant × the determinant. Considering the term $a_1^{n-1}a_2^{n-2}\dotsm a_{n-1}$ in the expansion of the general formula, we see this coefficient is $1$.
Let we consider the following problem: to find a third-degree polynomial $p(x)=c_3 x^3+c_2 x^2 + c_1 x + c_0$ that equals $1$ at $x=a_1$ and $0$ at $x=a_i$ with $i\neq 1$.
By Lagrange interpolation we know that the solution is given by $$ p(x) = \frac{(x-a_2)(x-a_3)(x-a_4)}{(a_1-a_2)(a_1-a_3)(a_1-a_4)} \tag{1}$$ so, for instance, $$ c_0 = p(0) = \frac{-a_2 a_3 a_4}{(a_1-a_2)(a_1-a_3)(a_1-a_4)}.\tag{2} $$ On the other hand, the problem can be stated in the following way: find $(c_0,c_1,c_2,c_3)\in\mathbb{R}^4$ such that $$ \begin{pmatrix}1 & a_1 & a_1^2 & a_1^3 \\ 1 & a_2 & a_2^2 & a_2^3 \\ 1 & a_3 & a_3^2 & a_3^3 \\ 1 & a_4 & a_4^2 & a_4^3 \end{pmatrix}\begin{pmatrix}c_0\\ c_1 \\ c_2 \\ c_3\end{pmatrix}=\begin{pmatrix}1\\ 0 \\ 0 \\ 0\end{pmatrix}\tag{3} $$ If we solve such problem through Cramer's rule we find: $$ c_0 = \frac{\det \begin{pmatrix}a_2 & a_2^2 & a_2^3 \\ a_3 & a_3^2 & a_3^3 \\ a_4 & a_4^2 & a_4^3 \end{pmatrix}}{\det \begin{pmatrix}1 & a_1 & a_1^2 & a_1^3 \\ 1 & a_2 & a_2^2 & a_2^3 \\ 1 & a_3 & a_3^2 & a_3^3 \\ 1 & a_4 & a_4^2 & a_4^3 \end{pmatrix}}=a_2 a_3 a_4\cdot \frac{\det \begin{pmatrix}1 & a_2 & a_2^2 \\ 1 & a_3 & a_3^2 \\ 1 & a_4 & a_4^2 \end{pmatrix}}{\det \begin{pmatrix}1 & a_1 & a_1^2 & a_1^3 \\ 1 & a_2 & a_2^2 & a_2^3 \\ 1 & a_3 & a_3^2 & a_3^3 \\ 1 & a_4 & a_4^2 & a_4^3 \end{pmatrix}}\tag{4}$$ and to prove the claim it is enough to compare $(2)$ and $(4)$, given: $$ \det \begin{pmatrix}1 & a_2 & a_2^2 \\ 1 & a_3 & a_3^2 \\ 1 & a_4 & a_4^2 \end{pmatrix}=\det \begin{pmatrix}1 & a_2 & a_2^2 \\ 0 & a_3-a_2 & a_3^2-a_2^2 \\ 0 & a_4-a_2 & a_4^2-a_2^2 \end{pmatrix}\\=(a_3-a_2)(a_4-a_2)\det\begin{pmatrix}1 & a_3+a_2 \\ 1 & a_4 + a_2\end{pmatrix}=(a_3-a_2)(a_4-a_2)(a_4-a_3).\tag{5} $$ As an alternative, you may directly use Gaussian elimination as performed in $(5)$ to tackle the original problem with a $4\times 4$ Vandermonde matrix.
But out of curiosity, what is the purpose of avoiding induction?