Let $\alpha$, $\beta \in \mathbb{R} \setminus \{0\}$.
Consider the matrix $$\begin{bmatrix} \frac{\alpha^2}{\beta^2}\mathbf{1}_{3 \times 3} & & & \\ & \frac{\alpha^2}{\beta^2}\mathbf{1}_{2 \times 2} \\ & & \frac{\alpha^2}{\beta^2}\mathbf{1}_{3 \times 3} \\ & & & \frac{\alpha^2}{\beta^2}\mathbf{1}_{2 \times 2} \\ \end{bmatrix} + \mathbf{I}_{10}$$ where $\mathbf{I}_{10}$ is the $10 \times 10$ identity matrix, $\mathbf{1}_{n \times n}$ denotes the $n \times n$ matrix of all ones, and all other entries are $0$ if not given above.
Is there an easy way to see that this matrix is positive definite?
I found this, which seems (if I'm not reading this incorrectly) to suggest that a necessary condition for $$\begin{bmatrix} \frac{\alpha^2}{\beta^2}\mathbf{1}_{3 \times 3} & & & \\ & \frac{\alpha^2}{\beta^2}\mathbf{1}_{2 \times 2} \\ & & \frac{\alpha^2}{\beta^2}\mathbf{1}_{3 \times 3} \\ & & & \frac{\alpha^2}{\beta^2}\mathbf{1}_{2 \times 2} \\ \end{bmatrix}$$ to be positive definite is that $$\begin{bmatrix} \frac{\alpha^2}{\beta^2}\mathbf{1}_{3 \times 3} & \\ & \frac{\alpha^2}{\beta^2}\mathbf{1}_{2 \times 2} \end{bmatrix}$$ is positive definite as well. But since $\mathbf{I}_{10}$ is added to the entire matrix, I'm not sure if this is useful at all.
A block diagonal matrix with square diagonal blocks is positive (semi)definite if each diagonal block is positive (semi)definite, as $$ \begin{bmatrix} u\\ v \end{bmatrix}^T \begin{bmatrix} A & 0\\ 0 & B\end{bmatrix}\begin{bmatrix} u\\ v \end{bmatrix} = u^T A u + v^T B v. $$ Note: The eigenvalues of such a block diagonal matrix are the eigenvalues of every diagonal blocks.
Here, each diagonal block is given by $A=I_n + r ee^T$ with $r\ge 0$ and $e = (1,\dotsc,1)^T \in \mathbb R^n$. For every $x\ne 0$ we have $$ x^T (I_n + r ee^T) x = \underbrace{x^T x}_{>0} + \underbrace{r (e^T x)^2}_{\ge 0} > 0. $$ Note: The eigenvalues of $A$ is $1$ and $1+rn$.