Determining if a symmetric matrix is positive definite

Solution 1:

Yes. Your matrix can be written as $$(a+b)I + a ee^T$$ where $I$ is the identity matrix and $e$ is the vector of ones. This is a sum of a symmetric positive definite (SPD) matrix and a symmetric positive semidefinite matrix. Hence it is SPD.

Solution 2:

In addition to Carl's answer (+1) you can use the Gershgorin circle theorem which is one I'll often try when I see diagonally dominant matrices like this one.

For each row, the sum of the absolute values of the off-diagonal entries is $2a$. The theorem then says that every eigenvalue is in a closed disc of radius $2a$ centered at the diagonal elements $2a + b$, so for any eigenvalue $\lambda$ we have $\lambda \geq b > 0$.