Symmetric matrix eigenvectors. I've noticed that if I have a square matrix of dimension n. Where the diagonal entries are all equivalent to one another and the off diagonal entries are also equivalent to one another (but not necessarily equivalent to the diagonals). Say: $$
\begin{bmatrix} 
a & b & b \\
b & a & b \\
b & b & a
\end{bmatrix} 
$$ 
I always get one eigenvector of the form 
$$
\begin{bmatrix} 
1 \\
1 \\
1
\end{bmatrix} 
$$ 
and the remainder with identical eigenvalues.
Is this a well known result? Can anyone direct me to a proof of such a result?
 A: It's well known and not particularly remarkable (it's common in linear algebra tests, and I recall one question on SE that I answered that involved precisely this type of a matrix).
Basically, write your matrix as a sum of a scaled identity and a projection:
$$A=(a-b)I+b \vec{n}\otimes\vec{n}$$
where $\vec{n}$ is a vector of all ones. This means there is only one preferred (special) direction in this system: $\vec{n}$. So $\vec{n}$ is the only vector that can have a special eigenvalue, and all the vectors from its orthogonal complement are equivalent.
Extract the eigenvalue of $\vec{n}$:
$$A\vec{n}=(a-b)\vec{n}+b\vec{n}(\vec{n}\cdot\vec{n})=(a+2b)\vec{n}$$
where $|\vec{n}|^2=3$ is the only place where the dimension of the matrix comes in.
The other two eigenvalues must then be $(a-b)$:
$$A\vec{n}_\perp=(a-b)\vec{n}_\perp+0=(a-b)\vec{n}_\perp$$
A: Consider the identity matrix $I$ and the matrix $J$ with a $1$ in every cell. Your matrix $A$ is now $A = bJ + (a-b)I$.
Multiplying a matrix by $b \neq 0$ does not change the eigenspaces, it just multiplies their corresponding eigenvalues by $b$.
Adding $(a-b)I$ to a matrix does not change the eigenspaces, it just increments their corresponding eigenvalues by $(a-b)$.
So the eigenspaces of $A$ are exactly the eigenspaces of $J$.
