Finding the eigenvalues of $A=\left(\begin{smallmatrix} a & 1 & 1 \\ 1 & a & 1 \\ 1 & 1 & a \\ \end{smallmatrix}\right)$ I would like to calculate the eigenvalues of the following matrix $A$, but the factorization of the characteristic polynomial does not seem to be easy to compute.
$A=\pmatrix{
a & 1 & 1 \\
1 & a & 1 \\
1 & 1 & a \\
},\ a\neq1,\ a\neq-2$
$f(\lambda)$ = Char$(A,\lambda)$ = $(a-\lambda)^3-3(a-\lambda)+2 = -\lambda^3 + 3a\lambda^2 + 3\lambda(1-3a^2) + (a-1)^2(a+2)$
I have thought about using the Rational-Root Theorem (RRT), so possible roots of $f(\lambda)$ are $(a-1),\ (-a+1),\ (a+2),\ (-a-2)$, and much more, as for example in the case $a=2$ we should also test whether $f(\pm2)=0$ or not, am I wrong?
The eigenvalues of $A$ are $a-1$ and $a+2$ (computed with Wolfram Alpha). This result can be obtained using RRT, computing $f(a-1)$ and $f(a+2)$ and realizing that both are equal to zero but, is there an easier and 'elegant' way to find these eigenvalues?
 A: One way to see $a+2$ is an eigenvalue is that $$A\begin{bmatrix}1\\1\\1\end{bmatrix}=\begin{bmatrix}a+2\\a+2\\a+2\end{bmatrix}.$$
Then you can use the fact that $x-(a+2)$ divides the characteristic polynomial.

More generally: if all the rows of $A$ add up to $\lambda$, then $\lambda$ is an eigenvalue.
A: Basically, you need to solve $(a-\lambda)^3-3(a-\lambda)+2 =0$ for $\lambda$. Don't expand the brackets, instead denote: $t=a-\lambda$. Then:
$$t^3-3t+2=0 \Rightarrow (t-1)^2(t+2)=0 \Rightarrow \\
t_1=1 \Rightarrow a-\lambda =1 \Rightarrow \lambda_1 =a-1\\
t_2=-2\Rightarrow a-\lambda =-2 \Rightarrow \lambda_2=a+2.$$
A: Hint: if $I$ denotes the identity matrix, then the eigenvalues of $A+cI$ are easily obtained from the eigenvalues of $A$:
$$
(A+cI)v=\lambda v
\iff
Av=(\lambda-c)v
$$
What if you take $c=1-a$?
A: i think so. the matrix in question is called the rank one perturbation of the identity matrix. that is $A = (a-1)I +uu^\top$ where $u$ is called the unit vector with all entries one. it is know that $uu^t$ has eigenvalues $uu^\top$  and zero with multiplicity dimension of $u$ - 1 and the associated eigenvectors $u$ and $u^\perp.$ the eigenvalues of $(a-1)I + uu^\top$ are $(a-1+u^top u = a+1$ and two fold $a-1$ the determinant is the product of these eigenvalues. that is $(a-1)^2(a+2).$
