Matrix reduction trigonalisaton

Question

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

Answer 1

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.