How to prove this matrix is rank n?

Question

Let $M = \begin{pmatrix} p_0 (1- p_0) & -p_0 p_1 &\ldots & p_0 p_n\\ -p_1 p_0 & p_1 (1-p_1) & \ldots & p_1 p_n\\ \vdots & & &\vdots\\ -p_np_0 &\ldots&&p_n(1-p_n)\end{pmatrix}$ where $\sum_i p_i = 1$ and $p_i >0$ for all i. How to show this matrix has rank n?

For clarity, $M_{i,j} = p_i(\delta_{ij} - p_j)$ and $\delta_{ij} = 1$ if $i = j$ and 0 otherwise, where $\sum_i p_i = 1$ and $p_i >0$.

I tried random values of $\{p_0, \ldots, p_n\}$ and it consistently returns n, so hoping for the rank be n.

Answer 1

For a start: if you add every other row to the first row you get a zero row because $$p_0(1-p_0)-p_1p_0-\cdots-p_np_0=p_0-p_0(p_0+p_1+\cdots+p_n)=0$$ in the first column, and similarly for all other columns. So the rank of the $(n+1)\times(n+1)$ matrix is at most $n$.

Sorry but I gotta go now. Hope somebody else can finish the problem by showing that the rank is not less than $n$.

Answer 2

Your matrix is $$M=D-uu^\top$$

, where $D=\text{diag}(p_0,p_1,\ldots,p_n)$ and $u=(p_0,p_1,\ldots,p_n)^\top$

Consequently, $$\operatorname{rank}(M)\ge |\text{rank}(D)-\operatorname{rank}(uu^\top)|=n+1-1=n$$

The matrix $M$ of order $(n+1)\times(n+1)$ is related to the dispersion matrix of a singular multinomial distribution ('singular' because of the restriction $\sum\limits_{k=0}^n p_k=1$). Hence rank of $M$ is at most $n$.

This is also seen from the fact that $$\det M=\left(\prod_{k=0}^n p_k\right)\left(1-\sum_{k=0}^n p_k\right)=0$$

So, $\operatorname{rank}(M)=n$.