Let $(x_1,...,x_n)$ be real numbers and M be an $n \times n$ matrix whose its column is given by the entries $x_i,x_i^2, x_i^3,...x_i^n$. Compute the determinant of M.

I computed the formula for the determinant of M in terms of $x_1...x_n$ but I wonder if I can find a specific value.


2 Answers 2


Hint: use this determinant formula


take $x_{n+1}=0$ and develop the Vandermonde $(n+1)\times (n+1)$-determinant relatively to the last line.


@Tsemo Aristide is absolutely correct, you can follow that link and also The Proof for your specific case here. However, this is a different kind of explanation for what you have, which is not a proof but I think it might help you better grasp the concept.

At the first glance, this looks very similar to the Vandermonde matrix; except we have

$$ M = \left[ {\begin{array}{ccccc} x_1 & x_2 & x_3 & \cdots & x_n \\x_1^2 & x_2^2 & x_3^2 &\cdots & x_n^2\\ \vdots & & \ddots & & \vdots \\ x_1 ^n & x_2^n & x_3^n & \cdots &x_n^n \\ \end{array} } \right]$$ Now let us consider the Vandermonde's matrix which is

$$V = \left[ {\begin{array}{ccccc} 1 & x_1 & x_1^2& \cdots & x_1^{n-1}\\ 1 & x_2 & x_2^2 &\cdots & x_2^{n-1}\\ \vdots & & \ddots & & \vdots \\ 1 & x_n& x_n & \cdots &x_n^{n-1}\\ \end{array} } \right]$$ As we can see, there are very closely related; so we want to find the matrix $X$ so that $$VX=M$$ and in that case the determinant of $M$ will be $$\det(M) = \det(V) \cdot \det(X)$$ Now let us look at an example of a $2 \times 2$ by $2 \times 2$ matrices, so assume that we have : $$\left[ {\begin{array}{cc} a & 0\\ 0 & b \\ \end{array} } \right] \times \left[ {\begin{array}{cc} 1 & 1\\ 2& 2 \\ \end{array} } \right] = \left[ {\begin{array}{cc} a & a\\ 2b& 2b \\ \end{array} } \right] $$ So we can see that by multiplying this diagonal matrix by a the square matrix, we can multiply each component of the first row by $a$ and second row by $b$. Now this is exactly what we want in order to go from our the Vandermonde matrix to the given matrix; furthermore this is very beneficial since we know the determinant of the diagonal matrix is the multiplication of the diagonals and we know the Vandermonde's determinant. So to get $M$ we will have

$VX = N $ where $$X = \left[ {\begin{array}{ccccc} x_1 & 0 & 0& \cdots & 0\\ 0 & x_2 & 0 &\cdots & 0\\ \vdots & & \ddots & & \vdots \\ 0&0 &0 & \cdots &x_n\\ \end{array} } \right] $$ Now $$VX = \left[ {\begin{array}{ccccc} x_1 & x_1^2 & x_1^3 & \cdots & x_1^n\\ x_2 & x_2^2 & x_2^2 &\cdots & x_2^n\\ \vdots & & \ddots & & \vdots \\ x_n & x_n^2 & x_n^3 & \cdots &x_n^n\\ \end{array} } \right] = N $$ This is almost what we want except we have the transpose of this matrix, so $(VX)^T = M$. Also, for a square matrix, the determinant of the transpose is the same as the determinant of the matrix, we will have : $$\det(X)\cdot\det(V) = \prod_{1\leq i \leq n}x_i \cdot \prod_{1 \leq i \leq j \leq n} (x_j - x_i) = \det(M^T) = \det(M)$$

Again this is not a proof, but I hope this clarified some stuff from both of the links.

  • $\begingroup$ Thank you @Ali Heydari that makes sense! $\endgroup$
    – Zu-fu Lin
    Oct 11, 2017 at 5:36
  • $\begingroup$ I believe a much simpler version would be: Assume $x_1 \neq 0$. Then take $x_1^i$ from the $i^{th}$ column to get the vandermond determinant with entries $\frac{x_2}{x_1} , \frac{x_3}{x_1}, ..., \frac{x_n}{x_1}$. And then solve for $x_1 = 0$ separately. $\endgroup$
    – Anon
    Apr 22, 2018 at 10:43

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