How do I find an invertible matrix $P$ such that $P^{-1}AP$ is upper triangular?
$$A=\begin{pmatrix}-3 & 1 & -1\\-7 & 5 & -1\\-6 & 6 & -2\end{pmatrix}$$
I can't really find a good general method for doing this online.
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Hi if you wanted a computational platform then wolfram Alpha can be of help. https://www.wolframalpha.com/input/?i=characteristc+polynomial+of+matrix+%7B%7B-3,1,-1%7D,%7B-7,5,-1%7D,%7B-6,6,-2%7D%7D
From that link we find that the Characteristic polynomial of that Matrix is having roots in the Field $\mathbf{R}$ so therefore it is triangulable in that Field.
It has Eigen values $4,-2,-2$ .
So the Eigen vector coresponding to 4 can be taken as first vector $v_1$ in the basis and then find a vector $v_2$ whose image under $A$ lies in the subspace spanned by $v_1$ and $v_2$ make sure they are linearly independent.
Then atlast take the characteristic vector of the eigen value 2 $v_3$ so you get a basis.
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That's not true. Take any Jordan block with real diagonal and the result you claim fails – José Alejandro Aburto Araneda Jan 01 '19 at 15:20
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Hi José Alejandro Aburto Araneda I'm stating the Result from Hoffman Kunze page 203.Theorem 5 . I don't fully understand your objection – Sundara Narasimhan Jan 01 '19 at 15:27
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@JoséAlejandroAburtoAraneda Hey refer to Hoffman kunze page 203,Theorem 5.Chapter 6 – Sundara Narasimhan Jan 01 '19 at 15:28
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The matrix $\begin{pmatrix} 1 & 1 \ 0 & 1\end{pmatrix}$ is not diagonalizable, and its characteristic polynomial have real roots. The theorem you are "using" talks about the minimal polynomial not the characteristic polynomial – José Alejandro Aburto Araneda Jan 01 '19 at 15:30