Finding the eigenvalue for a matrix I have the matrix of $A = 
\begin {pmatrix} 
2 && 1 && 1 \\ 
1 && 2 && 1 \\ 
1 && 1 && 2
\end {pmatrix}$ and am trying to find its eigenvalue.
My work so far:
$A-\lambda = \begin {pmatrix} 
2-\lambda && 1 && 1 \\ 
1 && 2-\lambda && 1 \\ 
1 && 1 && 2-\lambda
\end {pmatrix}$
Using laplace expansion:
$(2-\lambda)((2-\lambda)^2-1) - ((2-\lambda)-1) + (1-(2-\lambda))$
Using difference of squares:
$(2-\lambda)(((2-\lambda)-1)(((2-\lambda)+1)) - ((2-\lambda)-1) + (1-(2-\lambda))$
and this is where I got stuck. I can't get it to the point where I can set each point to $0$ to get the eigenvalues. Any help would be appreciated.
 A: Why don't you try opening the brackets as well? You have $$(2-\lambda)((2-\lambda)^2-1) - ((2-\lambda)-1) + (1-(2-\lambda))\\ = (2-\lambda)^3 - (2-\lambda) - (2-\lambda) + 1 + 1 - (2-\lambda)\\ = (2-\lambda)^3 - 3(2-\lambda) + 2$$
Take $\mu = 2 - \lambda$. You get $\mu^3 - 3\mu + 2$, and $\mu - 1$ is a divisor of the equation. $$\mu^3 - 3\mu + 2 = (\mu^2 + \mu - 2) (\mu - 1)=(\mu + 2)(\mu - 1)^2$$
Therefore $\mu = -2,1,1 \implies \lambda = 4,1,1$
A: Taking the determinant, we get $p(\lambda) = (2-\lambda)^3 + 2 - 3(2-\lambda)$ = $-\lambda^3 + 6\lambda^2 -12\lambda + 8 + 2 - 6 +3\lambda = \lambda^3+6\lambda^2-9\lambda + 4$. Factoring this, we get down to $(4-\lambda)(\lambda-1)^2$. So we have eigenvalues $\lambda_1 = 1$ with multiplicity 1 and $\lambda_2 = 1$ with multiplicity 2.
A nice trick for computing the determinant of a $3x3$ matrix is to:
1) rewrite the first two columns to the right of the matrix
2) multiply down the three diagonals, add these values
3) multiply up the three diagonals and subtract this value. This is equivalent to the determinant, i.e. for $\begin{bmatrix}a & b & c\\\ 
d&e&f\\\
g&h&i
\end{bmatrix}$, we obtain $aei+bfg+cdh-ceg-afh-bdi$. It makes factoring to find the eigenvalues a bit easier to identify. Hope this helps!
