How to deduce that A has an eigenvalue -1 with algebraic multiplicity 2 without calculating characterstic polynomial. $A = \begin{pmatrix}
0 & 1 &1 \\ 
1 & 0 &1 \\ 
1& 1 &0 
\end{pmatrix} $
The matrix $(A+I)$ has rank $1$ , so $-1$ is an eigenvalue with an algebraic multiplicity of at least $2$ .
I was reviewing my notes and I don't understand how the first statement implies the second one. 
Can anyone please explain how rank 1 of $(A + I)$ implies $-1$ is an eigenvalue with an algebraic multiplicity of $2$?
Thank you in advance.

It is well known that the algebraic multiplicity of an eigenvalue is greater or equal than the geometric multiplicity of that eigenvalue (i.e. the dimension of its eigenspace). Therefore, knowing that the rank of $(A-(-1)I)$ is $1$, you know that the algebraic multplicity of $-1$ is at least $2$ because if we denote the geometric multiplicity by $n$, $rank + n = 3$ in this case (dimension of the kernel + dimension of the image is $3$ for a $3 \times 3$ matrix). Therefore $n = 2$ which is the lower bound for the algebraic multiplicity.
EDIT : After reading Pierre-Yves's answer, I had a little flash : you can actually deduce more from this. Using the argument above gives you that the algebraic multiplicity of $-1$ is at least $2$, and you know that $2$ is an eigenvalue since $(1,1,1)$ is an eigenvector of $A$. The sum of the geometric multiplicities is $3$. Since $-1$ has a geometric multiplicity of two and $2$ has a geometric multiplicity of $1$ (it can't have more), we know that there are no other eigenvalues. (I never computed the polynomial! Yay =D)
Hope that helps,
