# 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?

-
It's a rank one update to a nonsingular matrix, so rank is at least $n-1$. –  hardmath Dec 12 '12 at 15:48

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 $\sum_j 1/x_j = 1$.
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.]