I need to diagonalise a matrix (or at least attempt to) but I am stuck on the factorisation of the eigenvalues.
The matrix goes as follow : $ \textbf{A} = \matrix{ 5-\Lambda & -1 & 2 \cr -1 & 5-\Lambda & 2 \cr 2 & 2 & 2-\Lambda \cr } $
To calculate the determinant, we have to do
$5-\Lambda * |\matrix{ 5-\Lambda & 2 \cr 2 & 2-\Lambda \cr } | + 1* |\matrix{-1 & 2 \cr 2 & 2-\Lambda \cr } | + 2*|\matrix{-1 & 2 \cr 5-\Lambda & 2 \cr }| $
Which gives me $5-\Lambda [ ( 5-\Lambda)(2-\Lambda)-4)+[-(2-\lambda)-4]+2[-2-2( 5-\Lambda)]$
Now I did a few steps and got to this point : $ (5-\Lambda)^2 (2- \lambda) - ( 5-\Lambda)*4 - (2-\lambda)-8-4( 5-\Lambda)$
Seeing that nothing good came out, I developped the second half of the equation to get $ (5-\Lambda)^2 (2- \lambda) - 20 + 4\lambda-2+\lambda-8-20+4\lambda)$
$ (5-\Lambda)^2 (2- \lambda) - 50+9\lambda$
Now i'm stuck..if I try to develop the left side I get
$(25-10\lambda + \lambda^2)(2-\lambda)- 50+9\lambda$
$50-20\lambda+2\lambda^2-25\lambda+10\lambda^2-\lambda^3-50+9\lambda$
$-100-36\lambda+12\lambda^2-\lambda^3$
Now I can see that something is up with the multiples of 6, but I am not too familiar with factorizing $^3$ Note that this is an old exam's question and it seems to me that this takes way too much time to do. Also, my teacher warned us to not develop the parenthesises when looking for $\lambda$, so i'm confused about what else I could of done.
As has been noted in the comments, you’ve made an algebraic error in calculating the characteristic polynomial. However, the eigenvalues of this matrix can be found without doing that.
First, observe that the sum of the columns is $(6,6,6)^T$, so $(1,1,1)^T$ is an eigenvector of $6$. Similarly, the sum of the first two columns minus twice the last is zero, so $(1,1,-2)^T$ is an eigenvector of $0$. The trace of the matrix is $12$, so the last eigenvalue is also $6$. It should be a simple matter to finish diagonalizing the matrix from here.