Generalised Diagonalisation

Let $\mathbb K$ be a field and $V$ be a finite-dimensional vector space over $V$ . Recall that $T : V \to V$ is diagonalisable if and only if it is similar to some simpler diagonal matrix $[T]_K : \mathbb K^n \to \mathbb K^n$ for some basis $K$ for $V$ . This condition is equivalent to the condition

$V = E_{\lambda_1} \oplus \cdots \oplus E_{\lambda_k},$

where $E_{\lambda_i} = \ker(T - \lambda I_V)$ . We abbreviate $T - \lambda \equiv T - \lambda I_V$ . If $c_T$ can be factored into (possibly repeated) linear factors, find constants $\alpha_i$ such that

$c_T(x) = (x-\lambda_1)^{\alpha_1}\cdot \cdots \cdot (x-\lambda_k)^{\alpha_k}.$

Then the minimal polynomial $m_T$ of $T$ can be factored into

$m_T(x) = (x-\lambda_1)^{\beta_1}\cdot \cdots \cdot (x-\lambda_k)^{\beta_k},\quad 1 \leq \beta_i \leq \alpha_i.$

Now, if $\mathbb K = \mathbb C$ , then the fundamental theorem of algebra guarantees this condition is satisfied. In this case, is $T$ similar to some sufficiently simple matrix $[T]_K$ for some basis $K$ of $V$ , even if $[T]_K$ is not diagonal?

One hint in this direction is that in this case, we can make the decomposition

$V = K_{\lambda_1} \oplus \cdots \oplus K_{\lambda_k},$

where $K_{\lambda_i} := \ker( ( T - \lambda_i )^{\beta_i} )$ is a generalised eigenspace (since the case $\beta_i = 1$ yields an actual eigenspace). That is, can we use this decomposition to derive a generalised diagonalisation?

Lemma 1. Define the matrix $\mathbf J : \mathbb K^n \to \mathbb K^n$ by $\mathbf J(\mathbf e_1) = \lambda \mathbf e_1$ and

$\mathbf J( \mathbf e_{i+1} ) = \mathbf e_i + \lambda \mathbf e_{i+1},\quad 1 \leq i \leq n-1.$

Then $(\mathbf J - \lambda)^n = \mathbf 0$ . Equivalently, $\ker((\mathbf J - \lambda)^n) = \mathbb K^n$ . Explicitly, we call

$\mathbf J \equiv \mathbf J_n(\lambda) := \begin{bmatrix} \lambda & 1 & 0 & \cdots & 0 \\ 0 & \ddots & \ddots & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & 0 \\ \vdots & & \ddots & \ddots & 1\\ 0 & \cdots & \cdots & 0 & \lambda \end{bmatrix} : \mathbb K^n \to \mathbb K^n$

a Jordan block of $\lambda$ with size $n$ .

Proof. The first condition yields $(\mathbf J - \lambda)(\mathbf e_1) = \mathbf 0$ . More generally, if $(\mathbf J - \lambda)^i (\mathbf e_i)= \mathbf 0$ , then

$\begin{aligned} (\mathbf J - \lambda)^{i+1}(\mathbf e_{i+1}) &= (\mathbf J - \lambda)^i ( ( \mathbf J - \lambda )( \mathbf e_{i+1}) ) \\ &= (\mathbf J - \lambda)^i (\mathbf e_i) \\ &= \mathbf 0. \end{aligned}$

Therefore, $(\mathbf J - \lambda)^n = \mathbf 0$ .

With some work, we can prove that $c_{\mathbf J}(x) = m_{\mathbf J}(x) = (x-\lambda)^n$ .

Lemma 2. Suppose there exist constants $\lambda_i \in \mathbb K$ and positive integers $\beta_i$ such that

$\mathbf J = \begin{bmatrix} \mathbf J_{\lambda_1}(\beta_1) & 0 & \cdots & 0 \\ 0 & \ddots & \ddots & \vdots \\ \vdots & \ddots & \ddots & 0 \\ 0 & \cdots & 0 & \mathbf J_{\lambda_k}(\beta_k)\end{bmatrix} : \mathbb K^n \to \mathbb K^n,$

where $n := \beta_1 + \cdots + \beta_k$ . Then for each $i$ and $j$ , $(\mathbf J - \lambda_i)^n (\mathbf e_j) = \mathbf 0$ . We call $\mathbf J$ a matrix in Jordan normal form.

Proof. If $1 \leq j \leq \beta_1$ , then

$\begin{aligned} (\mathbf J - \lambda_1)^n (\mathbf e_j) &= (\mathbf J_{\lambda_1} - \lambda_1)^n (\mathbf e_j) \\ &= (\mathbf J_{\lambda_1} - \lambda_1)^{\beta_1} (\mathbf e_j) = \mathbf 0.\end{aligned}$

More generally, if $\beta_i+1 \leq j \leq \beta_{j+1}$ , then

$\begin{aligned} (\mathbf J - \lambda_{i+1})^n (\mathbf e_j) &= (\mathbf J_{\lambda_{i+1}} - \lambda_{i+1})^n (\mathbf e_j) \\ &= (\mathbf J_{\lambda_{i+1}} - \lambda_{i+1})^{\beta_{i+1}} (\mathbf e_j) = \mathbf 0. \end{aligned}$

What we have studied is that if $c_{\mathbf A}$ can be factored into a product of linear factors, then there exist constants $\lambda_i \in \mathbb K$ and positive integers $\beta_i$ such that

$\mathbb K^n = K_{\lambda_1} \oplus \cdots \oplus K_{\lambda_k}.$

This yields the result $(\mathbf A - \lambda_i)^n(\mathbf e_j) = \mathbf 0$ for any $i, j$ . Can we go the other direction, just like diagonalisation? Yes!

Theorem 1 (Jordan Normal Form). If $c_T$ can be factored into a product of linear factors, then there exists a basis $B$ of $V$ such that $[T]_B$ is in Jordan normal form.

Proof Sketch. Suppose for each $\lambda_i$ , we can find a matrix $\mathbf J_i$ in Jordan normal form. Then the decomposition

$V = K_{\lambda_1} \oplus \cdots \oplus K_{\lambda_k}.$

yields a basis $B := B_1 \cup \cdots \cup B_k$ such that

$[T]_B = \begin{bmatrix} \mathbf J_1 & 0 & \cdots & 0 \\ 0 & \ddots & \ddots & \vdots \\ \vdots & \ddots & \ddots & 0 \\ 0 & \cdots & 0 & \mathbf J_k \end{bmatrix}.$

It remains to find a basis $B_i$ for $K_{\lambda_i}$ such that $[T|_{K_{\lambda_i}}]_{B_i}$ is in Jordan normal form, without loss of generality. Denote $S_i := T|_{K_{\lambda_i}} - \lambda_i$ and for any $\mathbf v \in V$ , let $\langle \mathbf v \rangle := \mathrm{span}\{S_i^j(\mathbf v) : j \in \mathbb N\}$ be the $S_i$ -cyclic subspace generated by $\mathbf v$ .

Fix $\mathbf v_i^{(1)} \in K_{\lambda_i}$ and find a basis $B_i^{(1)}$ for $\langle \mathbf v_i^{(1)} \rangle$ . If $\mathrm{span}(B_i^{(1)}) = K_{\lambda_i}$ , then we are done. Inductively, if $\mathrm{span}(B_i^{(1)} \cup \dots \cup B_i^{(j)}) \neq K_{\lambda_i}$ , find

$\mathbf v_i^{j+1} \in K_{\lambda_i} \backslash \mathrm{span}(B_i^{(1)} \cup \dots \cup B_i^{(j)})$

and obtain a basis $B_i^{(j+1)}$ for $\langle \mathbf v_i^{(j+1)} \rangle$ . Since $K_{\lambda_i}$ is finite-dimensional, this process must end, yielding the finite basis

$B_i = B_i^{(1)} \cup \dots \cup B_i^{(j_i)}$

of $K_{\lambda_i}$ , so that

$[T|_{K_{\lambda_i}}]_{B_1} = \begin{bmatrix} \mathbf J_{\lambda_i}(|B_i^{(1)}|) & 0 & \cdots & 0 \\ 0 & \ddots & \ddots & \vdots \\ \vdots & \ddots & \ddots & 0 \\ 0 & \cdots & 0 & \mathbf J_{\lambda_i}(|B_i^{(j_i)}|) \end{bmatrix} =: \mathbf J_i.$

I think this proof sketch more-or-less provides the key steps and intuition required to prove Theorem 1. Check out a more complete proof in this link, which effectively formalises our ideas using mathematical induction.

Essentially, any linear transformation $\mathbf A : \mathbb C^n \to \mathbb C^n$ can be split into subspaces that can be generated through the Jordan blocks—which are precise formulations of generalised diagonalisation.

Next time, we pivot to generalising right angles using inner products. In fact, we revisit a somewhat familiar idea—dot products.

—Joel Kindiak, 11 Mar 25, 1914H

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Generalised Diagonalisation

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