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Computing the determinant of a Vandermonde-like matrix

This is a problem from a students competition but it looks quite straightforward to solve.

The task is to compute the determinant of the following matrix.

$$A = \begin{bmatrix}1&2&3&...&n\\1^2&2^2&3^2&...&n^2\\...&...&...&...&...\\1^n&2^n&3^n&...&n^n\end{bmatrix}$$

What I noticed is that it looks similar to Vandermonde's matrix even though it's not quite that.

Let's pull out from its determinant a common multiple $k$ from the $k$-th column for each $k=1,2,3,...,n$.

We get

$$det(A) = n! \cdot \begin{vmatrix}1^0&2^0&3^0&...&n^0\\1^1&2^1&3^1&...&n^1\\...&...&...&...&...\\1^{n-1}&2^{n-1}&3^{n-1}&...&n^{n-1}\end{vmatrix}$$

The matrix whose determinant we got in the RHS of the last equality is the transpose of the Vandermonde matrix (with $\alpha_s = s$, for $s = 1,2,...,n$). Hence its determinant equals the determinant of the Vandermonde matrix itself (since we know that $det(A) = det(A^T)$).

So (leaving aside the $n!$ multiplier) the determinant we are left with now is Vandermonde's determinant for the numbers $1,2,3,...,n$. We know its value is equal to

$$\prod\limits_{1 \le i \lt j \le n} (j-i)$$

So the result we get now is

$$det(A) = n! \cdot \prod\limits_{1 \le i \lt j \le n} (j-i)$$

This is the same as

$$n! \cdot (n-1)^1 \cdot (n-2)^2 \cdot (n-3)^3\ ...\ 3^{n-3} \cdot 2^{n-2} \cdot 1^{n-1}$$

Finally it is not difficult to realize that this expression is equal to

$$n!\ \cdot (n-1)!\ \cdot (n-2)!\ ... \ 3! \cdot\ 2! \cdot\ 1!$$

So this is our final answer here for the determinant of $A$.