$6$ eigenvalues of a $4\times4$-matrix? I am struggling with determining the eigenvalues of the following (symmetric) matrix:
$$ A =\begin{pmatrix}
2 & 1 & 0 & 0 \\
1 & 2 & 0 & 0 \\
0 & 0 & 2 & 1 \\
0 & 0 & 1 & 2
\end{pmatrix} $$ 
What I did : $$ \operatorname{char}_X(A) =\det\begin{pmatrix}
2 -X & 1 & 0 & 0 \\
1 & 2 -X & 0 & 0 \\
0 & 0 & 2 -X & 1 \\
0 & 0 & 1 & 2 -X
\end{pmatrix} "=" \det\begin{pmatrix}
2 -X & 1 & 0 & 0 \\
0 & (2-X)^2-1 & 0 & 0 \\
0 & 0 & 2 -X & 1 \\
0 & 0 & 0 & (2-X)^2-1
\end{pmatrix}  = $$ $$=(2-X)^2 ((2-X)^2-1)^2 = 
(2-X)^2 (X-3)^2 (X-1)^2. $$
This yields $6$ eigenvalues. How could that happen? What is wrong here?

The asserted equality of two determinants is incorrect.  Let $M$ be the matrix whose determinant appears first; let $N$ be the matrix whose determinant appears after the "equals" sign.
You multiplied the second row by $2-X$.  That alters the determinant.  Then you added $-1$ times the first row to the second, and that does not alter the determinant.  Then you did the same in the third and fourth rows.  So you would have
$$
\det M = \frac 1 {(2-X)^2} \det N = (X-3)^2(X-1)^2.
$$
Another thing to be aware of is this:
$$
\det\begin{bmatrix} A & 0 \\ 0 & B\end{bmatrix} = \det A\cdot\det B.
$$
