Determining if a symmetric matrix is positive definite I have a symmetric matrix where all non-diagonal elements are positive and identical, and all diagonal elements are identical as well. For example, the $3 \times 3$ version of this matrix has the following form: 
$$  \left( \begin{array}{ccc}
2a+b & a & a \\
a & 2a+b & a \\
a & a & 2a+b \end{array} \right)  $$
Note that $a>0\ , b>0$. For such a simple form, is there an easy way of determining that the above matrix is positive definite in the general $n \times n$ 
 case? I'd like to show that the matrix is still positive definite when the dimension is higher. Thank you. 
 A: 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 $\lambda \geq b > 0$.
A: $$\begin{bmatrix} 
2a+b &    a &    a\\   
   a & 2a+b &    a\\
   a &    a & 2a+b\end{bmatrix} = (a+b) \, \mathrm I_3 + a \, 1_3 1_3^\top = (a+b) \, \mathrm I_3 + 3a \left(\frac{\, 1_3}{\sqrt{3}}\right) \left(\frac{\, 1_3}{\sqrt{3}}\right)^\top$$
Hence, the eigenvalues are $4a + b$ and $a+b$, with multiplicities $1$ and $2$, respectively. Since $a, b > 0$, both eigenvalues are positive, i.e., the given symmetric matrix is positive definite.
A: 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.
