Trigonalise a matrix

Question

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

• 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

Answer 1

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