Proof that a given matrix has rank $n-1$ 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?
 A: 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 $\sum_j 1/x_j = 1$.
A: Consider the matrix
$$A=\begin{pmatrix}1-x_1^{-1}&-x_2^{-1}&\ldots&-x_n^{-1}\\
-x_1^{-1}&1-x_2^{-1}&\ldots&-x_n^{-1}\\
\vdots&\vdots&\ddots&\vdots\\
-x_1^{-1}&-x_2^{-1}&\ldots&1-x_n^{-1}\end{pmatrix}$$
obtained by the one above dividing the $j$-th column by $x_j$.
Then, let $u=(1,1,\ldots,1)^t$, so that
$$Au=\begin{pmatrix}1-x_1^{-1}-x_2^{-1}-\ldots-x_n^{-1}\\
-x_1^{-1}+1-x_2^{-1}-\ldots-x_n^{-1}\\
\vdots\\
-x_1^{-1}-x_2^{-1}-\ldots+1-x_n^{-1}\end{pmatrix}=\begin{pmatrix}0\\0\\\vdots\\0\end{pmatrix}\;.$$
Hence, $u\in\ker A$, so $\mathrm{rk}\; A\leq n-1$.
On the other hand, set $D=\mathrm{diag}(x_1,\ldots, x_n)$ and $U=(u_{ij})$ with $u_{ij}=1$ for every $1\leq i,j\leq n$. Then $A+U=D$ and $\mathrm{rk}U=1$, $\mathrm{rk}D=n$, hence $\mathrm{rk} A\geq n-1$. [Edit: this is essentially the comment by hardmath and it's proved by noticing that $\mathbb{R}^n=\mathrm{Im}A+U\subseteq\mathrm{Im}A \oplus \mathrm{Im}U$ and, as $\mathrm{Im}U$ is one-dimensional, $\mathrm{Im}A$ has to be at least $n-1$-dimensional.]
