Editing Vandermonde matrix (section)

===Second proof: linear maps===
Let {{mvar|F}} be a [[field (mathematics)|field]] containing all <math>x_i,</math> and <math>P_n</math> the {{mvar|F}} [[vector space]] of the polynomials of degree less than or equal to {{mvar|n}} with coefficients in {{mvar|F}}. Let 
:<math>\varphi:P_n\to F^{n+1}</math>
be the [[linear map]] defined by 
:<math>p(x) \mapsto (p(x_0), p(x_1), \ldots, p(x_n))</math>.

The Vandermonde matrix is the matrix of <math>\varphi</math> with respect to the [[canonical basis|canonical bases]] of <math>P_n</math> and <math>F^{n+1}.</math>

[[Change of basis|Changing the basis]] of <math>P_n</math> amounts to multiplying the Vandermonde matrix by a change-of-basis matrix {{mvar|M}} (from the right). This does not change the determinant, if the determinant of {{mvar|M}} is {{val|1}}.

The polynomials <math>1</math>, <math>x-x_0</math>, <math>(x-x_0)(x-x_1)</math>, …, <math>(x-x_0) (x-x_1) \cdots (x-x_{n-1})</math> are [[monic polynomial|monic]] of respective degrees 0, 1, …, {{mvar|n}}. Their matrix on the [[monomial basis]] is an [[upper-triangular matrix]] {{mvar|U}} (if the monomials are ordered in increasing degrees), with all diagonal entries equal to one. This matrix is thus a change-of-basis matrix of determinant one. The matrix of <math>\varphi</math> on this new basis is 
:<math>\begin{bmatrix}
1 & 0 & 0 & \ldots & 0 \\ 
1 & x_1-x_0 & 0 & \ldots & 0 \\ 
1 & x_2-x_0 & (x_2-x_0)(x_2-x_1) & \ldots & 0 \\ 
\vdots & \vdots & \vdots & \ddots & \vdots \\ 
1 & x_n-x_0 & (x_n-x_0)(x_n-x_1) & \ldots &  
(x_n-x_0)(x_n-x_1)\cdots (x_n-x_{n-1})
\end{bmatrix}</math>.
Thus Vandermonde determinant equals the determinant of this matrix, which is the product of its diagonal entries. 

This proves the desired equality. Moreover, one gets the [[LU decomposition]] of {{mvar|V}} as <math>V=LU^{-1}</math>.