Let $x_1 \cdots x_n$ be real numbers $x_i>1$ such that
$$\frac{1}{x_1} + \cdots + \frac{1}{x_n} = 1$$
Is it true that the matrix
$$
\left[\begin{matrix}
x_1-1 & -1 & \cdots & -1 \\
-1 & x_2-1 & \cdots & -1 \\
\vdots & \vdots & \ddots & -1 \\
-1 & -1 & \cdots & x_n-1
\end{matrix}\right]
$$
has rank $n-1$? Is there a simple proof?
Best Answer
Yes, it's true. Here's a very elementary proof, using $\text{rk}(M) = n-\dim (\ker M)$. We have $M_{ij} = x_i\delta_{ij} - 1$, so $v \in \ker M$ becomes \begin{align*} &\sum_{j=1}^n (x_i\delta_{ij} - 1)v_j = 0 ~,~\forall~i\\ \Rightarrow& v_j = \frac{1}{x_j}\sum_{i=1}^n v_i~. \end{align*} So the kernel is one-dimensional (determined by the arbitrary constant $K = \sum_i v_i$). The constraint you get from summing the above over $j$ is satisfied identically due to your condition $\sum_j 1/x_j = 1$.