ASKSAGE: Sage Q&A Forum - RSS feedhttps://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Sun, 11 Dec 2022 10:08:28 +0100Jordan form and simultaneous diagonalizationhttps://ask.sagemath.org/question/65314/jordan-form-and-simultaneous-diagonalization/Let $m$ be a block diagonal matrix $diag(b_1, \dots, b_m)$. Let $(p_i)$ be the orthogonal projections such that $b_i = p_i m p_i$. If we compute the Jordan form of $m$ using SageMath as follows:
sage: jf, q = m.jordan_form(transformation=True)
**Question 1**: Does SageMath computes the Jordan form of $m$ by computing the Jordan form of each $b_i$?
(more precisely, $p_i$ commutes with $\pi^{-1} q \pi$, for some permutation matrix $\pi$?).
In fact, what we are really interested in here is the simultanenous diagonalization: let $m$, $n$ be two simultaneously diagonalizable matrix. Make the following:
sage: jf1, q1 = m.jordan_form(transformation=True)
sage: nn= ~q1 * n * q1
sage: jf2, q2 = nn.jordan_form(transformation=True)
**Question 2**: Is it true that the change-of-basis matrix $q=q_1q_2$ both diagonalizes $m$ and $n$?
A positive answer to Q1 should imply a positive answer to Q2.
If you know a better way to make simultaneous diagonalization, please let me know. More generally, we are also interested in simultaneous block-diagonalization, more precisely, if $m$ and $n$ does not commute, the $*$-algebra they generate is isomorphic to $\bigoplus_i M_{n_i}(\mathbb{C})$ with at least one $i$ such that $n_i>1$, and we are interested in the block-diagonalization according to this decomposition.Sébastien PalcouxSun, 11 Dec 2022 10:08:28 +0100https://ask.sagemath.org/question/65314/How does SageMath compute the Jordan form of a block diagonal matrix?https://ask.sagemath.org/question/65313/how-does-sagemath-compute-the-jordan-form-of-a-block-diagonal-matrix/ Let $M$ be a block diagonal matrix $diag(B_1, \dots, B_m)$. Let $(p_i)$ be the orthogonal projections such that $B_i = p_i M p_i$. If you compute the Jordan form of $M$ using SageMath as follows:
sage: jf, q = M.jordan_form(transformation=True)
**Question**: Is it true that $q$ commutes with $p_i$ for all $i$?
It would be true if SageMath computes the Jordan form of $M$ by computing the Jordan form of each $B_i$.Sébastien PalcouxSun, 11 Dec 2022 10:08:27 +0100https://ask.sagemath.org/question/65313/