How to find the eigenvalues and eigenvector without computation? The given matrix is 
$$
\begin{pmatrix}
2 & 2 & 2 \\
2 & 2 & 2 \\
2 & 2 & 2 \\
\end{pmatrix}
$$
so, how could i find the eigenvalues and eigenvector without computation?
Thank you
 A: For the eigenvalues, you can look at the matrix and extract some quick informations.
Notice that the matrix has rank one (all columns are the same), hence zero is an eigenvalue with algebraic multiplicity two. For the third eigenvalue, use the fact that the trace of the matrix equals the sum of all its eigenvalues; since $\lambda_1=\lambda_2=0$, you easily get $\lambda_3=6$.
For the eigenvectors corresponding to $\lambda=0$ sometimes it's not hard; in this case it's clear that $v_1=[1,-1,0]$ and $v_2=[0,1,-1]$ (to be normalized) are eigenvectors corresponding to $\lambda=0$. For the eigenvector corresponding to $\lambda_3$, it's not as obvious as before, and you might have to actually write down the system, finding $v_3=[1,1,1]$ (to be normalized).
A: Try to solve the following matrix equation, taking into account that your matrix is singular:
$$\begin{pmatrix}2&2&2\\2&2&2\\2&2&2\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}2(x+y+z)\\2(x+y+z)\\2(x+y+z)\end{pmatrix}=2\begin{pmatrix}x+y+z\\x+y+z\\x+y+z\end{pmatrix}=\lambda\begin{pmatrix}x\\y\\z\end{pmatrix}$$
A: If sum of all the elements in each rows or each columns are same then sum of elements in one row is nothing but one of the eigenvalue of that matrix.
    In above matrix sum of all the elements in each rows is 6.Hence 6 is the eigenvalue of that matrix.   
