LEC ALG4 9

Jordan canonical form

Let $V$ be a $k$-vector space, and $T\in {\text{End}}_k(V)$. The goal is to find a basis for $V$ in which the matrix of $T$ has a ‘nice’ form. We assume that $k$ is algebraically closed, but we only really require that the characteristic poly

Recall that to make a $\mathbb{Z}$-module $M$ an $R$-module, we need to specify a ring map $R\to {\text{End}}_{\mathbb{Z}}(M)$. Note that any ring is a $\mathbb{Z}$-algebra.

Now, let $V$ be a $\mathbb{Z}$-module. To make $V$ a $k$-vector space, we need to specify a ring map $\rho :k\to {\text{End}}_{\mathbb{Z}}(V)$. Since $k$ is a field, $\rho$ must be injective.

${\text{End}}_k(V)$, the set of $k$-linear maps from $V$ to $V$, is clearly a subset of ${\text{End}}_{\mathbb{Z}}(V)$. Recall that $T\in {\text{End}}_{\mathbb{Z}}(V)$ is $k$-linear iff $T(av)=aT(v)$ for all $a\in k$ and for all $v\in V$. This can be written as $T\circ \rho (a)=\rho (a)\circ T$ for all $a\in k$. Therefore, ${\text{End}}_k(V)$ is the centralizer of $\rho (k)$ in ${\text{End}}_{\mathbb{Z}}(V)$. Since $k$ is commutative, $\rho (k)$ lies inside the centralizer of $\rho (k)$, so $\rho (k)\subseteq {\text{End}}_k(V)$.

Next, let $k$ be a field, and $V$ a finite dimensional $k$-vector space. Consider the $k$-algebra $k[x]$. To make $V$ a $k[x]$ module, we need to specify a ring map $k[x]\to {\text{End}}_k(V)$.

Let $T\in {\text{End}}_k(V)$. The mapping $x\mapsto T$ gives us a map $k[x]\to {\text{End}}_k(V)$, with the scalars being mapped using $\rho$.

This makes $V$ into a $k[x]$-module, compatible with the $k$-vector space structure of $V$ when $k$ is considered to be a subring of $k[x]$. Thus, the basis that generates $V$ as a $k$-vector space will continue to generate $V$ as a $k[x]$-module, making $V$ a finitely generated $k[x]$-module.

Example 402.1(Example).

Suppose $V=k[x]/⟨x^2⟩$; by Exm 108.10, $V$ has a natural $k[x]$-module structure. What should the $T\in {\text{End}}_k(V)$ that $x\in k[x]$ gets mapped to be to realize this module structure as a map $k[x]\to {\text{End}}_k(V)$?

Firstly, as a $k$-vector space, $V\cong k^2$ with basis $\{\overline{1},\overline{x}\}$. In the natural $k[x]$-module structure, $x$ acts on these elements like so:

$$\begin{array}{rl}T(\overline{1})& =\overline{x}\\ T(\overline{x})& =\overline{0}.\end{array}$$

Thus, in the basis $\{\overline{1},\overline{x}\}$, $T$ is given by the matrix $\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]$.

Remark 402.2(Remark).

Suppose $V=V_1\oplus V_2$ as $k[x]$-modules, where $V_1$ and $V_2$ are $k[x]$-submodules of the $k[x]$-module $V$. Clearly, $V_1$ and $V_2$ are also $k$-submodules of the $k$-module $V$. The conditions $V_1\cap V_2=0$ and $V_1+V_2=V$, being properties of $V$ as an abelian group, are not impacted by the module structure on $V$. Thus, the equation $V=V_1\oplus V_2$ continues to hold in $k\text{-Mod}$.

The converse is not true. For example, $k[x]/⟨x^2⟩$, as a $k[x]$-module, cannot be written as a direct sum of two proper submodules¹. But as a $k$-vector space, it is isomorphic to $k^2$.

eddb40

Proposition 402.3(Proposition).

$V$ is a torsion $k[x]$-module.

Proof.

By Thm 391.9, write $V=V_f\oplus V_t$ as $k[x]$-modules, where $V_f$ is a free $k[x]$-module and $V_t$ is a torsion $k[x]$-module. By Rmk 2, $V=V_f\oplus V_t$ are $k$-vector spaces. Thus, we have ${\dim }_{k}V={\dim }_k{V}_f+{\dim }_k{V}_t$. If $V_f\ne 0$, then ${\dim }_k{V}_f$ is countably infinite, so ${\dim }_{k}V$ is countably infinite, a contradiction.□

Proof.

This proof isn’t ‘canon’, since we’d like to prove Caley-Hamilton using this development later.

Let $p(x)$ be the characteristic polynomial of $T$. Then, by C-H, $p(T)=0\in {\text{End}}_k(V)$. Thus, $p(x)\in k[x]$ annihilates $V$ as a $k[x]$-module.□

Using Thm 394.7 and Thm 398.12, we can now write

$$V\cong \frac{k[x]}{⟨p_1^{e_1}⟩}\oplus \cdots \oplus \frac{k[x]}{⟨p_s^{e_s}⟩},$$E1

as $k[x]$-modules, where the $p_i$‘s are irreducible polynomials (and not necessarily distinct) and $e_i⩾1$ for all $i$. Call the entity on the RHS $K$, and let $\phi :V\to K$ be an isomorphism. Using $\phi$, we can obtain subspaces $V_1,\dots ,V_s\subseteq V$ such that

$$V=V_1\oplus \cdots \oplus V_s.$$

Note that since $\phi$ is a $k[x]$-module homomorphism, multiplying by $x$ in $K$ retains the meaning of ‘applying $T$’:

We will implicitly identify elements of $K$ with their isomorphic counterparts in $V$.

Proposition 402.4(Proposition).

Suppose $V=V_1\oplus V_2$ as $k[x]$-modules. Let $B_i\subseteq V_i$ be $k$-basis for $i=1,2$. Then with respect to $B_1\bigsqcup B_2$, $T$ is given by a block diagonal matrix

where ${\ast }_1=T{|}_{V_1}$ and ${\ast }_2=T{|}_{V_2}$.

Proof.

Firstly, we have $V=V_1\oplus V_2$ as $k$-vector spaces, by Rmk 2. Next, since $V_1$ and $V_2$ are $k[x]$-submodules, we have $Tv=xv\in V_i$ when $v\in V_i$ for $i=1,2$. In other words, $V_1$ and $V_2$ are $T$-invariant subspaces of $V$. It follows that the matrix of $T$ is block diagonal in the basis $B_1\bigsqcup B_2$.□

Thus, if $B_1,\dots ,B_s$ are $k$-bases for the summands $k[x]/⟨p^{e_1}⟩,\dots ,k[x]/⟨p^{e_s}⟩$, $T$ is given by the matrix

Thus, by choosing our basis vectors from the invariant subspaces provided by (E1), it is possible to get the matrix of $T$ to be in block diagonal form. We now seek a specific choice of $B_1,\dots ,B_n$ such that the blocks themselves have a ‘nice’ form.

We now assume $k$ is algebraically closed, so all primes in $k[x]$ are of the from $x-\lambda$ for some $\lambda \in k$. The summands in (E1) now take the form

$$W_i=\frac{k[x]}{⟨(x-\lambda {)}^{e_i}⟩}.$$

An immediate candidate for $B_i$ is $\{\overline{1},\overline{x},\dots ,{\overline{x}}^{e_i-1}\}$. However, this is not the basis we are looking for.

Example 402.5(Example).

Suppose one of the summands in (E1) is of the form

$$W=\frac{k[x]}{⟨(x-\lambda {)}^2⟩}.$$

Consider the basis $B=\{\overline{1},\overline{x}\}$ ². In this basis, the matrix of $T$ (the action of which, remember, corresponds to multiplication by $x$) is given by

$$\begin{array}{r}\left[\begin{array}{cc}0& -{\lambda }^2\\ 1& 2\lambda \end{array}\right]\end{array}.$$

Although the difference isn’t very stark in the $2\times 2$ case, we seem to get a seemingly nicer matrix using the bases $\{\overline{1},\overline{x}-\lambda \}$:

$$\left[\begin{array}{cc}\lambda & 0\\ 1& \lambda \end{array}\right].$$

In general, the basis

$$\{w_1=\overline{1},w_2=\overline{(x-\lambda )},w_3=\overline{(x-\lambda {)}^2},\dots ,w_e=\overline{(x-\lambda {)}^{e-1}}\}$$E2

for $W$ of the form $k[x]/⟨(x-\lambda {)}^e⟩$ yields the matrix

$$\left[\begin{array}{ccccc}\lambda & & & & \\ 1& \lambda & & & \\ & 1& \lambda & & \\ & & \ddots & \ddots & \\ & & & 1& \lambda \end{array}\right]$$E3

with $\lambda$‘s on the diagonal and $1$‘s below the diagonal:

$$\begin{array}{rl}T{w}_i& =x\overline{(x-\lambda {)}^{i-1}}\\ & =\lambda \overline{(x-\lambda {)}^{i-1}}+(x-\lambda )\overline{(x-\lambda {)}^{i-1}}\\ & =\{\begin{array}{ll}\lambda w_i+w_{i+1}& i<e\\ \lambda w_i& i=e.\end{array}\end{array}$$

This is about as ‘nice’ as matrices get. Satisfied with our work, we make the definition

Definition 402.6(Jordan block).

A Jordan block of size $n$ and eigenvalue $\lambda$ is an $n\times n$ matrix of type (E3).

d71679

Definition 402.7(Jordan canonical form).

A matrix is said to be in Jordan canonical form if it is block diagonal with Jordan blocks on the diagonal.

$$\left[\begin{array}{cccc}{J}_1& & & \\ & J_2& & \\ & & \ddots & \\ & & & J_n\end{array}\right]$$

68d070

Theorem 402.8(Theorem).

Let $k$ be an algebraically closed field. Let $V$ be a finite dimensional $k$-vector space. Let $T:V\to V$ be $k$-linear. Then there exists a basis of $V$ with respect to which $T$ is given by a matrix in Jordan canonical form.

Proof.

The union of bases prescribed by (E2) for each $T$-invariant subspace of $V$ obtained in (E1) works.□

261664

Footnotes

1. $x^2(k[x]/⟨x^2⟩)=0$; use Thm 398.12. ↩

2. This is abuse of notation; when working in the wider context of $W$ being a subspace of $K$, ${\overline{1}}_W$ and ${\overline{x}}_W$ would be more accurate, albeit a bit clunky. ↩