Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How do you compute the semisimple part of the Jordan (SN) decomposition of a matrix $A\in M_n(K)$? The method I believe is correct is this:

  1. Compute the eigenvalues $a_1,\dots,a_r$ of $A$ and their generalized eigenspaces $\tilde{V}_{a_1},\dots,\tilde{V}_{a_r}$.
  2. Compute a basis $(b_{k,1},\dots,b_{k,d_k})$ for each generalized eigenspace $\tilde{V}_{a_k}$, and let $P$ be a matrix $(b_{1,1},\dots,b_{1,d_1},\dots,b_{k,1},\dots,b_{k,d_k},\dots,b_{r,1},\dots,b_{r,d_r})$, where each vector is regarded as a column vector.
  3. Let $B = a_1I_{d_1}\oplus\cdots\oplus a_rI_{d_r}$ and then $P^{-1}BP$ will be the semisimple part.

Is this a correct and convenient way to compute the semisimple part by hand?

share|improve this question
    
+1! Great question! –  Pierre-Yves Gaillard Jan 6 '12 at 13:18
add comment

2 Answers

up vote 2 down vote accepted

Let $p$ be the minimal polynomial of $A$. Then there is a unique polynomial $s\in K[X]$ of degree less than $\deg p$ such that $s(A)$ is the semi-simple part of $A$.

Assume, as we may, that $p$ splits in $K$ as, say, $$ p=\prod_{\lambda\in\Lambda}\ (X-\lambda)^{m(\lambda)} $$ with $m(\lambda) > 0$ for all $\lambda$.

Then $s$ is the unique degree less than $\deg p$ solution to the congruences $$ s\equiv\lambda\quad\bmod\quad(X-\lambda)^{m(\lambda)},\quad\lambda\in\Lambda, $$ and it is given by $$ s=\sum_{\lambda\in\Lambda}\ \lambda\ T_\lambda\left(\frac{(X-\lambda)^{m(\lambda)}}{p}\right)\frac{p}{(X-\lambda)^{m(\lambda)}}\quad, $$ where $T_\lambda(f)$ means "order less than $m(\lambda)$ Taylor polynomial of $f$ at $\lambda$".

EDIT. Here is a proof. Put $$ B_\lambda:=\frac{K[X]}{(X-\lambda)^{m(\lambda)}}\quad. $$

(A) We have canonical $K[X]$-algebra isomorphisms $$ K[A]\simeq\frac{K[X]}{(p)}\simeq\prod_{\lambda\in\Lambda}\ B_\lambda=:B, $$ the second isomorphism being given by the Chinese Remainder Theorem.

We way (and will) work in $B$ instead of working in $K[A]$.

Let $x\in B$ be the canonical image of $X$, and $e_\lambda$ the element of $B$ whose $\lambda$ component is $1$, and whose other components are $0$.

We must find the semi-simple part of $x$. But this is clearly the sum of the $\lambda e_\lambda$. In view of (A), this shows that, as claimed, $s$ is the unique degree less than $\deg p$ solution to the congruences $$ s\equiv\lambda\quad\bmod\quad(X-\lambda)^{m(\lambda)},\quad\lambda\in\Lambda, $$ and we're left with solving these congruences.

It's not harder to solve the general congruence system $$ s\equiv p_\lambda\quad\bmod\quad(X-\lambda)^{m(\lambda)},\quad\lambda\in\Lambda, $$ where the $p_\lambda\in K[X]$ are arbitrary.

The trick is to use the Ansatz $$ s:=\sum_{\lambda\in\Lambda}\ s_\lambda\ \frac{p}{(X-\lambda)^{m(\lambda)}}\quad,\quad\deg s_\lambda < m(\lambda), $$ which gives the solution $$ \sum_{\lambda\in\Lambda}\ T_\lambda\left(p_\lambda\ \frac{(X-\lambda)^{m(\lambda)}}{p}\right)\frac{p}{(X-\lambda)^{m(\lambda)}}\quad. $$

[Recall that $A$ admits a Jordan decomposition if and only if its eigenvalues are separable over $K$ (Bourbaki, Algèbre, Théorème VII.5.9.1). We assume here that such is the case.]

share|improve this answer
    
Could you give me more details about $T_\lambda$? (and what I really meant to ask was whether the method I described was correct.) –  Pteromys Jan 6 '12 at 11:47
1  
Dear @Pteromys: I'll write a proof as soon as possible, but I'm very slow. Thanks in advance for your patience. –  Pierre-Yves Gaillard Jan 6 '12 at 11:56
1  
Dear @Pteromys: I tried to give a proof. –  Pierre-Yves Gaillard Jan 6 '12 at 13:16
add comment

Your method is correct but requires to compute the root of the characteristic polynomial of $A$, which can't be done in general (there is no algorithm if $K=\mathbb{C}$, but if $K$ is finite there are algorithms such as Cantor-Zassenhaus).

The only way I know (but there may be other ones) to compute the semi-simple part of $A$ is the following : let $P(X) = \frac{\chi_A(X)}{GCD(\chi_A(X),\chi'_A(X))}$. Compute the sequence $(A_k)$ of matrices by $A_0=A$ and $A_{k+1} = A_k - P(A) P'(A)^{-1}$. Then for $k \geq \log_2 n$, $A_k$ is the semi-simple part of $A$ (the key point to prove this is to note that the semis-simple part of $A$ is a zero of $P(X)$).

share|improve this answer
    
Perhaps my wording was a bit confusing, but what I wanted to ask was all about computing it for matrices $\in M(\mathbb{R})$ by hand, e.g., when solving exercises on textbooks. –  Pteromys Jan 6 '12 at 14:20
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.