Let $A_n(x)$ be the $ n{\times}n$ matrix whose entries are $x^{\max(i,j)}$, find the max of $\det A_n(x)$ on the closed interval $[0,1]$

Question

I tried row reducing this matrix, and got that it is row equivalent to a lower triangular matrix of terms $(1-x)$. This would give determinant $(1-x)^n$, which in turn would imply the maximum of $\det A_n(x)$ attains a maximum at $x=0$? This is just a zero matrix?

Is this correct? In that case is the determinant for all other values of x less than $0$?

Thanks in advance for the help!

Answer 1

No, that's all wrong. For example, in the $2 \times 2$ case, $$\det \pmatrix{x & x^2\cr x^2 & x^2\cr} = x^3- x^4$$ which has its maximum at $x=3/4$.

Answer 2

A quick "sanity check": if we let $x = 0$, your formula gives the determinant as $1$. But the determinant of the zero matrix is $0$, so something has gone wrong.

Let's write it out: \begin{equation*} A_n(x) = \begin{pmatrix} x & x^2 & x^3 & x^4 & \dotsm \\ x^2 & x^2 & x^3 & x^4 & \dotsm \\ x^3 & x^3 & x^3 & x^4 & \dotsm \\ x^4 & x^4 & x^4 & x^4 & \dotsm \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix} \end{equation*} Now for the row reduction let's subtract $x$ times the row above it from each row, starting from the bottom. Then \begin{align*} \det A_n(x) &= \begin{vmatrix} x & \dotsm \\ 0 & x^2 - x^3 & \dotsm \\ 0 & 0 & x^3 - x^4 & \dotsm \\ 0 & 0 & 0 & x^4 - x^5 & \dotsm \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{vmatrix} \\ &= x^{n(n + 1)/2}(1 - x)^{n - 1} \end{align*} Does that help?