A Note on the Jordan Decomposition.

Abstract.
We show some kind of universal property for the Jordan decomposition of an endomorphism of a finite dimensional vector space.

This time, I want to share an observation on the Jordan decomposition, which is the main tool needed to show the existence of the Jordan normal form. Let me begin by introducing a more general notation, and show that the Jordan decomposition satisfies a kind of universal property.

Let $V$ be a vector space over a field $K$ and $\varphi : V \to V$ a linear map. We say that a subspace $W \subseteq V$ is $\varphi$ -invariant if $\varphi(W) \subseteq W$ . Another way to interpret this is to consider the $K$ -algebra $A = K[\varphi]$ ; then $V$ is an $A$ -module and the $A$ -submodules of $V$ are exactly the $\varphi$ -invariant subspaces of $V$ .

Definition.

An $A$ -decomposition of $V$ is a decomposition $V = \bigoplus_{i \in I} V_i$ of $A$ -submodules such that, for every $A$ -submodule $W$ of $V$ , one has $W = \bigoplus_{i \in I} (V_i \oplus W)$ .

Clearly, there always exists a trivial $A$ -decomposition of $V$ , namely $V$ itself. One can define a partial order on the set of $A$ -decompositions:

Definition.

Let $V = \bigoplus_{i \in I} V_i$ and $V = \bigoplus_{j \in J} W_j$ be two $V$ decompositions. We say that $\bigoplus_{i \in I} V_i \le \bigoplus_{j \in J} W_j$ if, for every $i \in I$ , there exists an $j \in J$ with $V_i \subseteq W_j$ .

Clearly, the trivial $A$ -decomposition is the maximum with respect to this order. One can ask whether a minimal $A$ -decomposition exists. In case it exists, it has a nice property:

Lemma.

Assume that $V = \bigoplus_{i \in I} V_i$ is a minimal $A$ -decomposition. Let $W = \bigoplus_{j \in J} W_j$ be another $A$ -decomposition. Then, for every $j \in J$ , there exists a subset $I_j \subseteq I$ with $W_j = \bigoplus_{i \in I_j} V_i$ .

Proof.

Define $I_j := \{ i \in I \mid V_i \subseteq W_j \}$ . Then $I = \bigcup_{j \in J} I_j$ is a disjoint union. Now, $\bigcup_{i \in I_j} U_i \subseteq W_j$ and $\bigoplus_{i \in I_j} U_i$ form a direct sum, whence $\bigoplus_{i \in I_j} U_i \subseteq W_j$ for all $j$ .

Now assume that $\bigoplus_{i \in I_j} U_i \subsetneqq W_j$ for some $j$ ; let $w \in W_j \setminus \bigoplus_{i \in I_j} U_i$ . Now, as $V = \bigoplus_{i \in I} V_i$ , we can write $w = \sum_{i \in I} v_i$ with $v_i \in V_i$ . Moreover, write $w = \sum_{t \in J} w_t$ with $w_t \in W_t$ . Clearly, we must have $w_t = \sum_{i \in I_t} v_i$ for every $t \in J$ . As $w \in W_j$ we have $w_t = 0$ for all $t \neq 0$ , whence $v_i = 0$ for all $i \not\in I_j$ . But this implies $w = \sum_{i \in I_j} v_i \in \bigoplus_{i \in I_j} V_i$ , a contradiction.

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Now one can ask when such a decomposition exists, and if it can be computed. An important case in which this is true is the one where $V$ is a finitely dimensional vector space over an algebraically closed field $K$ ; for example, $K = \C$ and $V = \C^n$ .

Definition.

Let $\lambda \in K$ . The generalized eigenspace of $\varphi$ with respect to $\lambda$ is $\displaystyle \GEig(\varphi, \lambda) := \{ v \in V \mid \exists n \in \N : (\varphi - \lambda \id)^n v = 0 \}.$

In case $\dim V = n < \infty$ , one has that $\GEig(\varphi, \lambda) = \ker (\varphi - \lambda)^n$ . Hence, generalized eigenspaces can be efficiently computed. Moreover, we have $\Eig(\varphi, \lambda) \subseteq \GEig(\varphi, \lambda)$ , and a simple argument shows that either both are trivial or both are non-trivial. Hence, the $\lambda \in K$ with $\GEig(\varphi, \lambda) \neq \{ 0 \}$ are exactly the zeroes of the characteristic polynomial of $\varphi$ .

Now note that $\varphi(\GEig(\varphi, \lambda)) \subseteq \GEig(\varphi, \lambda)$ . Hence, $\GEig(\varphi, \lambda)$ is $\varphi$ -invariant. We now have three lemmas:

Lemma.

Let $\lambda_1, \dots, \lambda_t$ be $t$ different eigenvalues of $\varphi$ . Then $\bigoplus_{i=1}^t \GEig(\varphi, \lambda_i)$ is a direct sum.

Proof.

Let $v_i \in \GEig(\varphi, \lambda_i$ with $\sum_{i=1}^t v_i = 0$ . We have to show that $v_i = 0$ for all $i$ . Assume that not all $v_i$ are zero, and that the relation is chosen minimal with respect to the number of nonzero $v_i$ .

Let $j \in \{ 1, \dots, t \}$ with $v_j \neq 0$ , and choose $n \in \N$ with $(\varphi - \lambda_j \id)^n v_j = 0$ . If $\psi := (\varphi - \lambda_j \id)^n$ , then $\sum_{i=1}^t \psi(v_i) = 0$ yields a second relation with $\psi(v_i) = 0$ . By minimality, we must have $\psi(v_j) = 0$ for all $j$ .

We will show that $(\varphi - \lambda_i \id)|_{\GEig(\varphi, \lambda)}$ is injective for $\lambda \neq \lambda_i$ , which gives $v_j = 0$ for all $j \neq 0$ and, therefore, $v_i = 0$ , a contradiction.

Let $v \in \GEig(\varphi, \lambda)$ with $(\varphi - \lambda_i \id) v = 0$ . Assume that $v \neq 0$ and let $n \in \N$ be maximal with $w := (\varphi - \lambda \id)^n v \neq 0$ ; in that case, $\varphi(w) = \lambda w$ and $\displaystyle (\varphi - \lambda_i) w = (\varphi - \lambda_i) (\varphi - \lambda)^n v = (\varphi - \lambda)^n (\varphi - \lambda_i) v = (\varphi - \lambda)^n 0 = 0,$ whence we get $\lambda_i w = \varphi(w) = \lambda w$ , which is only possible for $w = 0$ , a contradiction. Hence, we must have $v = 0$ , i.e. $\varphi - \lambda_i \id$ is injective on $\GEig(\varphi, \lambda)$ .

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Lemma.

Assume that $\dim V < \infty$ and let $\lambda \in K$ . Then there exists an $\varphi$ -invariant subspace $W \subseteq V$ such that $V = W \oplus \GEig(\varphi, \lambda)$ .

Proof.

Set $\psi := \varphi - \lambda \id$ . ConsiderS the chains $\displaystyle \{ 0 \} \subseteq \ker \psi \subseteq \ker \psi^2 \subseteq \ker \psi^3 \subseteq \dots$ and $\displaystyle V \supseteq \image \psi \supseteq \image \psi^2 \supseteq \image \psi^3 \supseteq \dots$ Clearly, there exists an $s \in \N$ with $\image \psi^s = \image \psi^{s+1}$ as $\dim V < \infty$ . Now one easily shows $\image \psi^s = \image \psi^{s+i}$ for all $i \in \N$ . By the Dimension Formula, we have $\dim \ker \psi^{s+i} ={} & \dim V - \dim \image \psi^{s+i} \\ {}={ } & \dim V - \dim \image \psi^s = \dim \ker \psi^s$ for all $i \in \N$ , whence $\ker \psi^{s+i} = \ker \psi^s$ for all $i \in \N$ . But then $\GEig(\varphi, \lambda) = \ker \psi^s$ and $\dim \GEig(\varphi, \lambda) + \dim \image \psi^s = \dim V$ .

Set $W := \image \psi^s$ and let $w \in W$ , i.e. let $v \in V$ with $\psi^s(v) = w$ . Then $\displaystyle \varphi(w) = \varphi (\varphi - \lambda \id)^s v = (\varphi - \lambda \id)^s \varphi(v) \in \image \psi^s,$ whence $W$ is $\varphi$ -invariant. Now it suffices to show that $W \cap \GEig(\varphi, \lambda) = 0$ . Now $\image \psi^{s+1} = \image \psi^s$ , whence for every $w \in W$ there exists some $v \in \image \psi^s$ with $\psi(v) = w$ . But this means that $\psi|_W$ is surjective, whence $\ker(\psi|_W) = \{ 0 \}$ . But then $\ker \psi^s \cap W = \ker (\psi^s|_W) = \ker (\psi|_W)^s = \{ 0 \}$ .

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Lemma.

Assume that $\dim V < \infty$ and that the characteristic polynomial $\chi_\varphi$ of $\varphi$ splits into linear factors. Let $\lambda_1, \dots, \lambda_t$ be all eigenvalues of $\varphi$ . Then $V = \bigoplus_{i=1}^t \GEig(\varphi, \lambda)$ .

Proof.

We proceed by induction on $\dim V$ . For $\dim V = 0$ this is cleary. Hence, assume $\dim V \ge 1$ and let $\lambda$ be an eigenvalue of $\varphi$ . Choose an $\varphi$ -invariant subspace $W$ with $V = \GEig(\varphi, \lambda) \oplus W$ . We have $\dim \GEig(\varphi, \lambda) \ge \dim \Eig(\varphi, \lambda) \ge 1$ , whence $\dim W < \dim V$ . Now $\displaystyle \chi_\varphi = \chi_{\varphi|_W} \cdot \chi_{\varphi|_{\GEig(\varphi, \lambda)}},$ whence the characteristic polynomial of $\varphi|_W$ splits into linear factors as well.

Let $\lambda'_1, \dots, \lambda'_s$ be the eigenvalues of $\varphi|_W$ . Then, by induction, we have $W = \bigoplus_{i=1}^s \GEig(\varphi|_W, \lambda_i')$ . Now $\GEig(\varphi|_W, \lambda_i') = W \cap \GEig(\varphi, \lambda_i')$ , whence $W \subseteq \bigoplus_{i=1}^s \GEig(\varphi, \lambda_i')$ .

Finally, note that $\lambda \neq \lambda_i'$ for all $i$ , as this would contradict $W \cap \GEig(\varphi, \lambda) = \{ 0 \}$ . Therefore, $V = \GEig(\varphi, \lambda) \oplus \bigoplus_{i=1}^s \GEig(\varphi, \lambda_i')$ . Moreover, we must have $\{ \lambda_1, \dots, \lambda_t \} = \{ \lambda, \lambda'_1, \dots, \lambda_s' \}$ as the dimensions of the generalized eigenspaces for all $\lambda_i$ must be non-zero, whence “ $\supseteq$ ” must hold. The converse holds because every non-trival generalized eigenvector yields a non-trivial eigenvector to the same value.

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Therefore, we get:

Corollary (Jordan Decomposition).

Let $K$ be algebraically closed and assume that $\dim V < \infty$ . Then, for every endomorphism $\varphi$ of $V$ , there exist $\lambda_1, \dots, \lambda_t \in K$ such that $\displaystyle V = \bigoplus_{i=1}^t \GEig(\varphi, \lambda_i)$ is an $K[\varphi]$ -decomposition.
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Proof.

We have to show that this yields an $K[\varphi]$ -decomposition. For that, let $W$ be a $\varphi$ -invariant subspace of $V$ . Consider $\varphi|_W$ ; this is an endomorphism of $W$ whose set of eigenvalues is a subset of the set of eigenvalues of $\varphi$ . Hence, by the previous lemma applied to $\varphi|_W$ , we have $\displaystyle W = \bigoplus_{i=1}^t \GEig(\varphi|_W, \lambda_i) = \bigoplus_{i=1}^t (\GEig(\varphi, \lambda_i) \cap W),$ what we had to show.

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We can now prove our main result, namely that the generalized eigenspace decomposition is exactly the minimal $K[\varphi]$ -decomposition of $V$ :

Theorem.

Let $K$ be algebraically closed and $\dim V < \infty$ . Then the minimal $K[\varphi]$ -decomposition of $V$ is given by $\displaystyle V = \bigoplus_{\lambda \in K} \GEig(\varphi, \lambda).$

Note that we do not need that $K$ is algebraically cloesd, but only that $\chi_\varphi$ splits over $K$ .

Proof.

Let $V = \bigoplus_{i=1}^n V_i$ be a $K[\varphi]$ -decomposition. Assume that there exists some $\lambda \in K$ and $1 \le i < j \le n$ such that $V_i \cap W \neq \{ 0 \} \neq V_j \cap W$ with $W := \GEig(\varphi, \lambda)$ ; if this would not exist, we would have $\bigoplus_{\lambda \in K} \GEig(\varphi, \lambda) \le \bigoplus_{i=1}^n V_i$ .

Assume that we can find eigenvectors $v \in V_i \cap W$ and $w \in V_j \cap W$ . Then $\varphi(v + w) = \lambda (v + w)$ , whence $v + w$ is an eigenvector as well. But then $W' := \langle v + w \rangle$ is an $\varphi$ -invariant subspace of $V$ with $W' \subseteq V_i \oplus V_j$ , but $(W' \cap V_i) \oplus (W' \cap V_j) = \{ 0 \} \subsetneqq W'$ , a contradiction that $\bigoplus_{i=1}^n V_i$ is a $K[\varphi]$ -decomposition.

We now show that $W \cap V_i$ contains an eigenvector. As $V_i$ is $\varphi$ -invariant, we can consider $\psi := \varphi|_{V_i}$ . Now $\GEig(\psi, \lambda) = W \cap V_i \neq \{ 0 \}$ , whence we must have $\Eig(\psi, \lambda) \neq \{ 0 \}$ . Hence, there exists some $v \in W$ , $v \neq 0$ with $\varphi(v) = \psi(v) = \lambda v$ .

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Tags: Jordan decomposition, universal property

This entry was posted on Tuesday, May 5th, 2009 at 01:33 Europe/Berlin and is filed under Linear Algebra. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

A Note on the Jordan Decomposition.

Abstract.

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