Trigonalise a matrix 
Let $\mathbf{A}=\begin{bmatrix} 
    0 & 1 & 1  \\ 
   -1 & 1 & 1  \\
   -1 & 1 & 2
  \end{bmatrix}$
  
  
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*Trigonalise a matrix
  

Could someone trigonalise this matrix in details way, i wanna see the difference between yours and the french one  
to start here's  $P_{1}(\lambda)=(\lambda -1)^3$
note that some authors in book's frnech use :
$$T = \left( \begin{array}{ccc}  \lambda_{1} & 1 & 0 \\  0 & \lambda_{2} & 1 \\  0 & 0 & \lambda_{3} \\ \end{array} \right)$$
and other use $$T = \left( \begin{array}{ccc}  \lambda_{1} & a & b \\  0 & \lambda_{2} & c \\  0 & 0 & \lambda_{3} \\ \end{array} \right)$$
im wondering wich way that you using in usa english's books

From @Ian's comment below, the author appears to be saying to "make triangular by similarity transformations", which gives something of the form (or like you added to the comment).
$$T = \left( \begin{array}{ccc}  \lambda_{1} & a & b \\  0 & \lambda_{2} & c \\  0 & 0 & \lambda_{3} \\ \end{array} \right)$$
We try two things: $1.)~$ Diagonalize when possible, and if that fails, $2.)~$ Find the Jordan Normal Form.
For this specific problem, we have:
$$\mathbf{A}=\begin{bmatrix} 
    0 & 1 & 1  \\ 
   -1 & 1 & 1  \\
   -1 & 1 & 2
  \end{bmatrix}$$
We find the characteristic polynomial and eigenvalues using $|A - \lambda I| = 0$, resulting in:
$$-\lambda ^3+3 \lambda ^2-3 \lambda +1 = 0 \implies -(-1 + \lambda)^3 = 0 \implies \lambda_{1, 2, 3} = 1$$
Next, we want to find three linearly independent eigenvectors using $[A- \lambda I]v_i = 0$, but this is not always possible due to algebraic and geometric difference (deficient matrices), so we have to resort to generalized eigenvectors.
So, we have $[A - \lambda_1 I]v_1 = [A -I]v_1 = 0$ as:
$$\left(
\begin{array}{ccc}
 -1 & 1 & 1 \\
 -1 & 0 & 1 \\
 -1 & 1 & 1 \\
\end{array}
\right)v_1 = 0$$
If we put that in row-reduced-echelon-form (RREF), we arrive at:
$$\left(
\begin{array}{ccc}
 1 & 0 & -1 \\
 0 & 1 & 0 \\
 0 & 0 & 0 \\
\end{array}
\right) v_1 = 0$$
This only provides one linearly independent eigenvector as:
$$v_1 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}$$
Now, we need to find two more generalized eigenvectors and there are many approaches to that. One approach is to try $[A - \lambda I]v_2 = v_1$, yielding an augmented RREF of:
$$\begin{pmatrix} 1 & 0 & -1 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}$$
This yields $a = c, b = 1$, so choose $c = 0$, thus:
$$v_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$$
Repeating this process, we set up and solve $[A - I]v_3 = v_2$, yielding an augmented RREF of:
$$\begin{pmatrix} 1 & 0 & -1 & -1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 0 \end{pmatrix}$$
This yields $a = -1 + c, b = -1$, so choose $c = 0$, thus:
$$v_3 = \begin{pmatrix} -1 \\ -1 \\ 0 \end{pmatrix}$$
We can now form $P = (v_1~|~v_2~|~v_3)$ as:
$$P = \begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 1 & 0 & 0 \end{pmatrix}$$
This yields the Jordan Normal Form:
$$T = PAP^{-1} = \left( \begin{array}{ccc}  1 & 1 & 0 \\  0 & 1 & 1 \\  0 & 0 & 1 \\ \end{array} \right)$$
Note: there are various ways to solve these problems.


*

*You want to look for references with the Jordan Normal Form.

*You can sometimes infer $T$ (see references below).


Some sources for generalized eigenvectors and the Jordan Normal Form are:


*

*Wiki

*Notes 1

*Notes 2

*Jordan Normal Form and Chaining

*Book - Abstract Algebra by Dummit and Foote

*Books - many books on linear algebra

