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?

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any help would be appreciated


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:


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.
  • $\begingroup$ i think the question was trigonalisation and they said that A can be semblable of trigonalisation matrix and to achieve that they chooce the best form of matrix is jordan form because it's by default trigonalisation and then they can choose $C_{k}$ such that $AC_k=C_k+C_{k-1}$ due to the form of jordan contian the 1 up of its diagonal am i right ? $\endgroup$ – Educ Dec 28 '14 at 14:04
  • $\begingroup$ yes does mean similar $\endgroup$ – Educ Dec 28 '14 at 14:19
  • $\begingroup$ i will ask question about trigonalisation and let me see your way of how to trigonalise $\endgroup$ – Educ Dec 28 '14 at 15:02
  • $\begingroup$ could you add some names of good books in english wich treat this as details example or some web site homeroke with solution $\endgroup$ – Educ Dec 28 '14 at 18:32
  • $\begingroup$ please if you find my questions useful, Please recall to upvote and/or indicate that are intersting . $\endgroup$ – Educ Dec 28 '14 at 18:45

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