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Homeproof of determinant of the Vandermonde matrix

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# proof of determinant of the Vandermonde matrix

To begin, note that the determinant of the $n\times n$ Vandermonde matrix (which we shall denote as ‘$\Delta$’) is a homogeneous polynomial of order $n(n-1)/2$ because every term in the determinant is, up to sign, the product of a zeroth power of some variable times the first power of some other variable , $\ldots$, the $n-1$-st power of some variable and $0+1+\cdots+(n-1)=n(n-1)/2$.

Next, note that if $a_{i}=a_{j}$ with $i\neq j$, then $\Delta=0$ because two columns of the matrix would be equal. Since $\Delta$ is a polynomial, this implies that $a_{i}-a_{j}$ is a factor of $\Delta$. Hence,

$\Delta=C\prod_{{1\leq i<j\leq n}}(a_{j}-a_{i})$ |

where C is some polynomial. However, since both $\Delta$ and the product on the right hand side have the same degree, $C$ must have degree zero, i.e. $C$ must be a constant. So all that remains is the determine the value of this constant.

One way to determine this constant is to look at the coefficient of the leading diagonal, $\prod_{n}(a_{n})^{{n-1}}$. Since it equals 1 in both the determinant and the product, we conclude that $C=1$, hence

$\Delta=\prod_{{1\leq i<j\leq n}}(a_{j}-a_{i}).$ |

## Mathematics Subject Classification

15A57*no label found*65F99

*no label found*65T50

*no label found*

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