Linear Algebra, eigenvalues and eigenvectors For any $m \times m$ matrix, I will get a characteristic  polynomial of degree $m$
with $m$ eigenvalues.
But for the matrix 
$$A = \pmatrix{2 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 & 0 & 1}$$
I got the characteristic polynomial
$$P(A)=t(t+1)(1-t)^3.$$
This means $5$ eigenvalues: $\{1,1,1,-1,0\}$.
Did I do some thing wrong?
 A: I calculated the same polynomial and I got 
$$P(X)= X^2 (X-1)^3 (X-4) \,.$$
Note that $tr(A)=7$ has to be the sum of eigenvalues.

Just to get you started:
$$\det(A-tI)= \det \pmatrix{2-t & 1 & 1 & 1 & 1 & 1 \\ 1 & 1-t & 0 & 1 & 0 & 1 \\ 1 & 0 & 1-t & 0 & 0 & 1 \\ 1 & 0 & 0 & 1-t & 0 & 0 \\ 1 & 0 & 0 & 0 & 1-t & 0 \\ 1 & 0 & 0 & 0 & 0 & 1-t}$$
Subtract the 6th row from 4th and 5th:
$$\det(A-tI)= \det \pmatrix{2-t & 1 & 1 & 1 & 1 & 1 \\ 1 & 1-t & 0 & 1 & 0 & 1 \\ 1 & 0 & 1-t & 0 & 0 & 1 \\ 0 & 0 & 0 & 1-t & 0 & t-1 \\ 0 & 0 & 0 & 0 & 1-t & t-1 \\ 1 & 0 & 0 & 0 & 0 & 1-t}$$
Now, $(1-t)$ common factor on rows 4 and 5.
$$\det(A-tI)= (t-1)^2\det \pmatrix{2-t & 1 & 1 & 1 & 1 & 1 \\ 1 & 1-t & 0 & 1 & 0 & 1 \\ 1 & 0 & 1-t & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 0 & 1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 1-t}$$
Next add Column 4 and column 5 to Column 6, and you can get a smaller $4 \times 4$ determinant....
A: In this particular case, it's not hard to find the eigenvalues and eigenvectors without computing the characteristic polynomial.  Write out the equations for the entries of
 $A v - \lambda v = 0$:
$$
\eqalign{2\,v_{{1}} &+v_{{2}}+v_{{3}}+v_{{4}}+v_{{5}}+v_{{6}}-\lambda\,v_{{1}}=0\cr v_{{1}} &+v_{{2}}+v_{{4}}+v_{{6}}-\lambda\,v_{{2}}=0\cr
v_{{1}}&+v_{{3}}+v_
{{6}}-\lambda\,v_{{3}}=0\cr
v_{{1}}&+v_{{4}}-\lambda\,v_{{4}}=0\cr v_{{1}}&+v_
{{5}}-\lambda\,v_{{5}}=0\cr v_{{1}}&+v_{{6}}-\lambda\,v_{{6}}=0\cr}$$
From the last three equations we see that (unless $\lambda = 1$) $v_4 = v_5 = v_6 = v_1/(\lambda-1)$.
The second and third equations then become
$$ \eqalign{\frac{1+\lambda}{\lambda-1} v_1 - (\lambda - 1) v_2 &= 0\cr
            \frac{\lambda}{\lambda-1} v_1 - (\lambda - 1) v_3 &= 0\cr}$$
so $v_2 = (1+\lambda)/(\lambda-1)^2$ and $v_3 = \lambda v_1/(\lambda-1)^2$.
The first equation then becomes
$$ \frac{\lambda^2 (4-\lambda)}{(\lambda-1)^2} v_1 = 0$$
so (since $v_1 = 0$ would make all $v_i = 0$, which we don't want), $\lambda = 0$ or $4$, 
and taking $v_1 = 1$ gives us the eigenvectors
$$ \pmatrix{1\cr 5/9 \cr 4/9 \cr 1/3 \cr 1/3 \cr 1/3\cr} \ \text{for} \ \lambda = 4,
\pmatrix{1 \cr 1\cr 0 \cr -1\cr -1\cr -1\cr} \ \text{for} \ \lambda=0 $$
We still need to consider the case $\lambda = -1$, in which the equations become
$$ \eqalign{ v_1 &+ v_2 + v_3 + v_4 + v_5 + v_6 = 0\cr
             v_1 &+ v_4 + v_6 = 0\cr
             v_1 &+ v_6 = 0\cr
             v_1 &= 0\cr}$$
so $v_1 = v_4 = v_6 = 0$ and $v_5 = -v_2 - v_3$ with $v_2, v_3$ arbitrary.  This gives us eigenvectors
$$ \pmatrix{0 \cr 1 \cr 0 \cr 0\cr -1 \cr 0\cr}, \ \pmatrix{0 \cr 0 \cr 1 \cr 0\cr -1 \cr 0\cr}  \ \text{for}\ \lambda=1 $$
Note, by the way, that there are only four linearly independent eigenvectors: eigenvalue $0$ has algebraic multiplicity $2$ and geometric multiplicity $1$, while eigenvalue $1$
has algebraic multiplicity $3$ and geometric multiplicity $2$.
