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.Thu, 02 Feb 2023 04:29:22 +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.Sun, 11 Dec 2022 10:08:28 +0100https://ask.sagemath.org/question/65314/jordan-form-and-simultaneous-diagonalization/Comment by Sébastien Palcoux for <p>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:</p>
<pre><code>sage: jf, q = m.jordan_form(transformation=True)
</code></pre>
<p><strong>Question 1</strong>: 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 $\pi$?).</p>
<p>In fact, what we are really interested in here is the simultanenous diagonalization: let $m$, $n$ be two simultaneously diagonalizable matrix. Make the following: </p>
<pre><code>sage: jf1, q1 = m.jordan_form(transformation=True)
sage: nn= ~q1 * n * q1
sage: jf2, q2 = nn.jordan_form(transformation=True)
</code></pre>
<p><strong>Question 2</strong>: Is it true that the change-of-basis matrix $q=q_1q_2$ both diagonalizes $m$ and $n$? </p>
<p>A positive answer to Q1 should imply a positive answer to Q2.</p>
<p>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.</p>
https://ask.sagemath.org/question/65314/jordan-form-and-simultaneous-diagonalization/?comment=66184#post-id-66184More informations is available at https://ask.sagemath.org/question/65752/who-is-the-encoder-of-jordan_form-in-sagemathThu, 02 Feb 2023 04:29:22 +0100https://ask.sagemath.org/question/65314/jordan-form-and-simultaneous-diagonalization/?comment=66184#post-id-66184