How to calculate eigenvalues of this matrix? I am trying to solve the following equation to get the eigenvalues of this (n + 1 $\times$ n + 1) matrix m:
$$m = 
\begin{bmatrix}
a & a1^T \\
a1 & bJ
\end{bmatrix}$$
Where:
$$
a, b \in R, 1^T = [1.....1]^T
$$ and J is a (n $\times$ n) matrix of all $1$s. 
so far, I  have done the following(I is the identity matrix here):
$$det(m) = 
\begin{bmatrix}
a - \lambda & a1^T \\
a1 & bJ - \lambda I
\end{bmatrix} = 0 \\
abJ - a\lambda I - \lambda bJ + \lambda^2 I - a1^Ta1 = 0 \\ \lambda^2I - \lambda(aI + bJ) + abJ - na^2 = 0
$$
because
$$a1^Ta1 = [a....a][a....a]^T = na^2$$
How do I proceed from here?
 A: Your computation of the determinant is not correct: you are treating $J$ as if it were a scalar, when it is a matrix. It should be obvious since you get a degree two equation on $\lambda$ instead of degree $n$. 
For this matrix, it is easier to look at the eigenvalues directly. I will assume $a\ne b$ and both nonzero, since it is the non-trivial case. 
The equation $mx=\lambda x$ gives the equations 
$$\tag1
a\sum_{j=1}^n x_j=\lambda x_1,\ \ ax_1+b\sum_{j=2}^n x_j=\lambda x_2=\cdots=\lambda x_n. 
$$
Consider first the case $\lambda=0$. Now the equations are 
$$
\sum_{j=1}^n x_j=0,\ \ ax_1+b\sum_{j=2}^n x_j=0.
$$
Solving for $x_1$ in the first equation we get $(b-a)\sum_{j=2}^n x_j=0$. Then $x_1=-\sum_{j=2}^n x_j=0$. So the eigenspace for $\lambda=0$ is the subspace
$$
\{x:\ x_1=0,\ \sum_{j=2}^n x_j=0\},
$$
which has dimension $n-2$. 
Now consider the case $\lambda\ne0$. We obtain from $(1)$ that $x_2=\ldots=x_n$. They cannot be zero, because $(1)$ would become $ax_1=\lambda x_1$ and $ax_1=0$, a contradiction. So assume that $x_2=1$. Now $(1)$ becomes 
$$
(n-1)=(\lambda-a)x_1, \ \ \ ax_1+b(n-1)=\lambda;
$$
solving for $\lambda$ we obtain a quadratic equation on $\lambda$ that will give us the other two eigenvalues. Solving the quadratic quickly gives me 
$$
\lambda =\frac{a+b(n-1)\pm\sqrt{[b(n-1)+a]^2-4(n-1)(b-1)a}}{2}
$$
