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 clear that $A+I$ is diagonalizable with $(0,0,3)$ on the diagonal. 
Thus, $A$ is diagonalizable with $(-1,-1,2)$ on the diagonal. 
Justification of the first claim:
A vector is in the kernel of $A+I$ if and only if the sum of its coordinates is zero. 
The vector $(1,1,1)$ is an eigenvector for $A+I$ with eigenvalue $3$.
EDIT. We can apply the following observation to $B:=A+I$:
If $B$ is a rank one $n$ by $n$ matrix with entries in a field, then 
$\bullet$ either the trace of $B$ is zero, and $B$ is similar to the direct sum of $(\begin{smallmatrix}0&1\\ 0&0\end{smallmatrix})$ and a zero matrix, 
$\bullet$ or the trace $t$ of $B$ is nonzero, and $B$ is similar to the direct sum of the scalar $t$ and a zero matrix. 
Indeed, either the kernel (which is a hyperplane) contains the image (which is a line), or it doesn't.
