How to derive the Vandermonde Determinant?

Question

I watched this video https://www.youtube.com/watch?v=87iJTcXqTKY explaning the Vandermonde Determinant I understood everything but I was wondering why the guy never mentioned the (-1)^(i+j) term used in solving determinants?

Answer 1

Step 1: Argue that the determinant of the Vandermonde matrix is a polynomial of degree $n-1$ in $x_1$. This is argued by considering cofactor expansion. If one were to actually compute the the determinant using cofactor expansion, there would be a ton of $(-1)^{i+j}$'s here. However, instead of doing the computation, the video performs a "thought experiment" to determine the shape of the determinant. And concludes that it is a polynomial of degree $n-1$ in $x_1$.

Interlude: The formula for cofactor expansion would be a mess to compute. The goal of the video is figure out the determinant without going through the very long calculation of cofactor expansion.

Step 2: Determine that $x_1=x_2,\cdots,x_1=x_n$ are roots of the determinant. This is done without actually computing the determinant, but, instead using the properties of the determinant. If one had fully calculated the determinant in Step 1 and plugged in $x_1=x_2$, then the result would be zero, but since the video didn't actually compute the determinant, you must use other means to figure this out. Since a polynomial of degree $n-1$ has $n-1$ roots (counted with multiplicities), this means that the determinant is of the form $$ g(x_2,\cdots,x_n)(x_1-x_2)(x_1-x_3)\cdots (x_1-x_n). $$ Here $g$ is some polynomial that does not depend on $x_1$.

Step 3: Figure out $g(x_2,\cdots,x_n)$. Since $g(x_2,\cdots,x_n)$ is the coefficient of $x_1^{n-1}$, this can be calculated using cofactor expansion. And can be seen to be another Vandermonde matrix.