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.

—Joel Kindiak, 11 Mar 25, 1914H

Response

The Spectral Theorems – KindiakMath

September 8, 2025 at 10:34 pm

[…] operators are diagonalisable (though all operators over will have a Jordan normal form, details here). But diagonalisable operators are convenient since we can compute with relative ease. […]

LikeLike

Generalised Diagonalisation

Share this:

Response

Leave a comment Cancel reply