Matrix reduction trigonalisaton 
Let 
  $
  \mathbf{A}=\begin{bmatrix} 
    2 & -1 & -1  \\ 
    2 & 1 &  -2\\
    3 & -1 & -2
  \end{bmatrix}
   $
  
  
*
  
*Trigonalise a matrix
  

in process of trigonalisation of matrix in question they choose $AC_3=C_3+C_2$ I'm wondering why exactly and what is algorithm behind that?

any help would be appreciated 
 A: They are finding the Jordan Normal Form of the matrix $A$. 
They need to find three linearly independent eigenvectors for the three eigenvalues. 
Due to the repeated eigenvalue, they need to find a Generalized Eigenvector for  $\lambda = 1$ due to the geometric and algebriac multiplicity. 
This can be written as $(A- \lambda I)C_3 = (A-I)C_3 = C_2$, hence $AC_3 = C_3 + C_2$.
$P$ is comprised of the three linearly independent eigenvectors as column vectors that correspond to each of the eigenvalues, that is:
$$P = ( C_1 ~~ C_2 ~~ C_3 )$$
You can see examples at:


*

*Finding the Jordan canonical form of this upper triangule $3\times3$ matrix

*Triple Root

*Many others on Math Stack Exchange


Update
It appears that the author meant "make triangular by similarity transformations", for example:
$$T = \left( \begin{array}{ccc}  \lambda_{1} & a & b \\  0 & \lambda_{2} & c \\  0 & 0 & \lambda_{3} \\ \end{array} \right)$$
In this case, you can (in this order):


*

*$1.)~$ Try to diagonalize the matrix, which meets the definition. If that fails, 

*$2.)~$ Find the Jordan Normal Form as I describe above and the author shows.

