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I have the following question:
For the following matrix $A$, find an invertible matrix $P$ over $\mathbb{C}$ such that $P^{-1}AP$ is upper triangular: \begin{equation} A= \pmatrix{ 1 & 1 & 1 \\ 1 & 1 & 1 \\ 0 & 0 & 0} \end{equation} I have calculated the characteristic polynomial: $ \lambda^2(2-\lambda) $. From that I deduced that we have two eigenvalues:

  • $\lambda=0$ with algebraic multiplicity $2$.
  • $\lambda=2$ with algebraic multiplicity $1$.

For $\lambda=0$ the eigenvectors are of the form: \begin{equation} \alpha \pmatrix{-1\\0\\1} + \beta \pmatrix{-1\\1\\0}, \end{equation} where $\alpha$ and $\beta$ constants.
Hence, the eigenspace is two-dimensional, i.e the geometric multiplicity is $2$.
Similarly, for $\lambda=2$ we get: \begin{equation} \gamma \pmatrix{1\\1\\0} \end{equation}
where $\gamma$ a constant.
Hence, the geometric multiplicity of $\lambda=2$ is $1$.

Now the JCF form of $A$ is: $J_1(0) \oplus J_1(0) \oplus J_1(2)$.
This is the part I get confused. The JCF gives me the matrix:$\pmatrix{0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 2}$, which I have checked is what I get if I use $P= \pmatrix{-1 & -1 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 0}$ in $P^{-1}AP$. This is the diagonal form not the upper triangular form.

I know for a fact that using for e.g. $P= \pmatrix{1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1}$ will result in indeed an upper triangular matrix, namely $\pmatrix{2 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0}$.

It is to my understanding that that if we take the eigenvectors as columns of $P$ then computing the $P^{-1}AP$ will result in an upper triangular matrix. What exactly is the process I should follow to produce an upper triangular matrix? Is the process I am doing above not used for triangularisation? How are Jordan matrices involved in triangularisation?
I have also tried to find the generalised eigenvectors using $(A−\lambda I)^2 v=0$ , $(A−\lambda I)^3 v=0$ etc. but due to the nature of $A$ I am not getting any solutions. I am a bit lost on how to deduce the matrix $P$.

(A note: I have not done Schur decomposition and is not in my course's syllabus.)

Thank you in advance for any solutions/suggestions.

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    $\begingroup$ "This is the diagonal form, not the upper triangular form." The question is asking you to make $P^{-1} A P$ an upper triangular matrix. If you end up with a diagonal matrix then that is perfectly valid; there is not a unique answer. $\endgroup$ – angryavian Aug 17 '18 at 18:06
  • $\begingroup$ You do not necessarily need JNF for trigonalizing your matrix, see the answers to this question. $\endgroup$ – Dietrich Burde Aug 17 '18 at 18:10
  • $\begingroup$ @DietrichBurde Thank you for your comment. I have seen that question. The answer provided uses Schur Decomposition which is not the way I am supposed to follow. Upon further inspection of its comments I have tried to find the generalised eigenvectors using $(A−\lambda I)^2v=0$ , $(A−\lambda I)^3 v=0$ but due to the specific nature of $A$ I don't seem to be getting any solutions. Thoughts on what I should use instead for P? $\endgroup$ – Nick Aug 17 '18 at 18:30
  • $\begingroup$ Did you see my answer there? Since it is only a $3\times3$-matrix, we can easily solve the $9$ linear equations from $SA=AS$ in the entries, so that $SAS^{-1}$ is upper-triangular. $\endgroup$ – Dietrich Burde Aug 17 '18 at 18:38
  • $\begingroup$ Possible duplicate of How do I find upper triangular form of a given 3 by 3 matrix?? $\endgroup$ – Sil Aug 18 '18 at 6:55

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