Determinant of symmetric matrix $(A-\lambda I)$ If we have a matrix $(A-\lambda I)$ which is:
$\left(
\begin{array}{ccc}
1-\lambda & -1 & 2 \\ 
-1 & 1-\lambda & 2 \\ 
2 & 2 & 2-\lambda \\ 
\end{array}
\right)
$
Then it's determinant can be written as : $(-1)^n(\lambda-\lambda_1)(\lambda-\lambda_2)(\lambda-\lambda_3)$. In this case what will $\lambda_1$,$\lambda_2$ and $\lambda_3$ be equal to? And how do we determine it's value given that the matrix is symmetric?
 A: $\left( \begin{array}{ccc}
1-\lambda & -1 & 2 \\ 
-1 & 1-\lambda & 2 \\ 
2 & 2 & 2-\lambda \\ 
\end{array} \right)
$ = $\left( \begin{array}{ccc}
2-\lambda & -2+\lambda & 0 \\ 
-1 & 1-\lambda & 2 \\ 
2 & 2 & 2-\lambda \\ 
\end{array} \right)
=(-2+\lambda)
\left( \begin{array}{ccc}
-1 & 1 & 0 \\ 
-1 & 1-\lambda & 2 \\ 
2 & 2 & 2-\lambda \\ 
\end{array} \right) =
(-2+\lambda)
\left( \begin{array}{ccc}
-1 & 0 & 0 \\ 
-1 & -\lambda & 2 \\ 
2 & 4 & 2-\lambda \\ 
\end{array} \right)= (-1)(-2+\lambda)
\left( \begin{array}{ccc}
 -\lambda & 2 \\ 
4 & 2-\lambda \\ 
\end{array} \right)
= (-1)(-2+\lambda)
\left( \begin{array}{ccc}
4 -\lambda & 4 -\lambda \\ 
4 & 2-\lambda \\ 
\end{array} \right)= 
(-2+\lambda)(\lambda-4)
\left( \begin{array}{ccc}
1 & 1\\ 
4 & 2-\lambda \\ 
\end{array} \right)
= 
(-2+\lambda)(\lambda-4)
\left( \begin{array}{ccc}
0 & 1\\ 
2+\lambda & 2-\lambda \\ 
\end{array} \right)=(-2+\lambda)(\lambda-4)(2-\lambda)=(-1)(\lambda-2)(\lambda-4)(\lambda-2)
$ Therefore, 
$\lambda_1=2,\lambda_2=2,\lambda_3=4$.
A: You could simply expand the determinant, which is not that much work for a $3\times 3$.
Or you could set $\lambda=0$ and guess the eigenvalues of the resulting matrix, for example $(1\,{-1}\,0)^T$ and $(1\,1\,1)^T$ are obvious eigenvectors of eigenvalue $2$. The fact that the matrix is symmetric guarantees us that all eigenvalues will be real.
A: HINT : Solve $\det(A-\lambda I)=0$, with:
$$\begin{vmatrix}1-\lambda&-1&2\\-1&1-\lambda&2\\2&2&2-\lambda\end{vmatrix}=(1-\lambda)\begin{vmatrix}1-\lambda&2\\2&2-\lambda\end{vmatrix}-(-1)\begin{vmatrix}-1&2\\2&2-\lambda\end{vmatrix}+2\begin{vmatrix}-1&1-\lambda\\2&2\end{vmatrix}$$
