Is the following matrix positive definite?

Question

Is there an easy way to prove / disprove this?

$$ (X)_{ij} = \begin{cases} \dfrac{k}{n}& \text{if}\ i = j \\ \dfrac{k(k-1)}{n(n-1)} & \text{otherwise} \\ \end{cases} $$

where $X \in \mathbb{R}^{n \times n}$ and $1 < k < n$

Answer 1

It's positive definite because it is in the form of $aI+bee^T$, where $e=(1,1,\ldots,1)^T$ and $a,b$ are some positive scalars.

Answer 2

You can also do it by direct computation,

Suppose $\mathbf{x} := (x_1,\dots,x_n) \in \mathbb{R}^n$, with $\mathbf{x} \neq \mathbf{0} \in \mathbb{R}^n.$ Then the $i^{\text{th}}$ component of $X\mathbf{x}$, which we write as $(X\mathbf{x})_i$ is given by \begin{align}(X\mathbf{x})_i &= \frac kn x_i + \sum_{j \neq i} \frac{k(k-1)}{n(n-1)}x_j \\ &= \frac kn \left(x_i + \frac{k-1}{n-1}\sum_{j \neq i} x_j\right). \end{align} So \begin{align}\mathbf{x}^T X \mathbf{x} &= \sum_{i=1}^n x_i (X \mathbf{x})_i \\ &= \frac kn \sum_{i=1}^n\left( x_i^2 + \frac{k-1}{n-1} \sum_{j \neq i}x_i x_j \right) \\ &= \frac kn \left(\sum_{i=1}^n x_i^2 + \frac{k-1}{n-1}\sum_{i=1}^n \sum_{j \neq i} x_i x_j \right). \end{align}

In the case that $\sum_{i=1}^n \sum_{j \neq i} x_i x_j \geq 0,$ we easily get $\mathbf{x}^T X \mathbf{x}>0$.

However, when $\sum_{i=1}^n \sum_{j \neq i} x_i x_j < 0, $ we have \begin{align} \frac kn \left(\sum_{i=1}^n x_i^2 + \frac{k-1}{n-1}\sum_{i=1}^n \sum_{j \neq i} x_i x_j \right) &> \frac kn \left(\sum_{i=1}^n x_i^2 + \sum_{i=1}^n \sum_{j \neq i} x_i x_j \right)\\ &= \frac kn \left(\sum_{i=1}^n x_i \right)^2 \\ &> 0. \end{align} So in either case, we have $\mathbf{x}^T X \mathbf{x} >0$. Hence, $X$ is positive definite.