Eigenvalue of $A-aI_3$ Question: Let $A=\begin{pmatrix}  a+1 & 1 & 1 \\ 1 & a+1 & 1 \\  1 & 1 & a+1\end{pmatrix}$. Show that $A-aI_3$ has eigenvalue of 3. Also find eigenvector.
My thinking: 
I know that we have to apply characteristic polynomial $|A- \lambda I|$ to find the eigenvalue. I don't understand the part $A-aI_3$. What should I suppose to do here with this $A-aI_3$ ? I am kinda confused. Your kind suggestion will be appreciated. 
 A: $A-aI_3=\begin{pmatrix}  1 & 1 & 1 \\ 1 & 1 & 1 \\  1 & 1 & 1\end{pmatrix}$.
What is $\begin{pmatrix}  1 & 1 & 1 \\ 1 & 1 & 1 \\  1 & 1 & 1\end{pmatrix}\begin{pmatrix}  1  \\ 1  \\  1 \end{pmatrix}?$
A: Alternative answer:  $A-aI_3=\begin{pmatrix}  1 & 1 & 1 \\ 1 & 1 & 1 \\  1 & 1 & 1\end{pmatrix}$ clearly has eigenvalue $0$ 
with independent eigenvectors $\begin{pmatrix}  1  \\ -1  \\  0 \end{pmatrix}$ and $\begin{pmatrix}  1  \\ 0  \\  -1 \end{pmatrix}.$
The sum of the eigenvalues is the trace of the matrix, which is $3$, so the other eigenvalue is $3$.  
An eigenvector corresponding to the eigenvalue $3$ is in the kernel of  $\begin{pmatrix}  -2 & 1 & 1 \\ 1 & -2 & 1 \\  1 & 1 & -2\end{pmatrix}$ , i.e. $(-2,1,1)\times(1,-2,1)=(-3,-3,-3)$.
A: Notice that $A-aI$ is the matrix with just ones. So $(A-aI)(1,1,1) = 3(1,1,1)$
A: Another ad hoc alternative:
Note that $B=A-aI = e e^T$, where$e$ is a vector of ones.
To solve $B v = t v$, we note that ${\cal R B} = \operatorname{sp} \{ e \}$, so
the only non zero eigenvalue of $B$ can be found by solving
$e e^T e = t e$, so get $t=e^T e = 3$.
Since $\dim {R B} = 1$, we see that $\dim \ker B = 2$ and so $B$ has two zero eigenvalues (meaning there are two linearly independent $v_1,v_2$ such that
$B v_1 = 0, B v_2 = 0$.
Since $\ker B = \ker e^T$, we can find two suitable $v_k$ by solving $e^T v = 0$.
