8 Chapter 8. When Diagonalization Fails: Jordan Canonical Form

Generalized eigenvectors, nilpotent motion, and the hidden structure of repeated eigenvalues

Guiding question.
If a matrix does not have enough eigenvectors to be diagonalized, what is the next best coordinate system?

In Chapter 7, diagonalization told a beautiful story. If a matrix has enough eigenvectors, then we can choose an eigenvector basis and the matrix becomes diagonal:

\[ A=PDP^{-1}. \]

In that coordinate system, every coordinate evolves independently. Powers, exponentials, and dynamical systems become easy.

But not every matrix has enough eigenvectors. A repeated eigenvalue may give only one eigendirection. At first, this looks like failure. Jordan canonical form explains that the failure is structured. Missing eigenvectors are replaced by generalized eigenvectors, and these generalized eigenvectors form chains. Each chain becomes one Jordan block.

Main idea

Jordan form is diagonalization plus correction terms for missing eigenvectors:

\[ \boxed{A=PJP^{-1},\qquad J=\text{block diagonal matrix of Jordan blocks}.} \]

A diagonal matrix is the special case in which every Jordan block has size $1$.

AI and coding companion

Ask an AI tool to explain the difference between an eigenvector and a generalized eigenvector using a simple $2\times2$ Jordan block. Then verify the formulas by hand and with Python.

8.1 8.1 The obstruction: not enough eigenvectors

Consider

\[ A=\begin{bmatrix}2&1\\0&2\end{bmatrix}. \]

The only eigenvalue is $\lambda=2$ because

\[ \det(A-\lambda I)=(2-\lambda)^2. \]

But

\[ A-2I=\begin{bmatrix}0&1\\0&0\end{bmatrix}, \]

\[ \ker(A-2I)=\left\{\begin{bmatrix}x\\0\end{bmatrix}:x\in\mathbb R\right\}. \]

Thus the eigenspace is one-dimensional, even though the algebraic multiplicity of $2$ is two. We do not have enough eigenvectors to diagonalize $A$.

Warning

A repeated eigenvalue does not automatically cause a problem. The problem occurs when the eigenspace is too small.

Proof: why this matrix is not diagonalizable

A $2\times2$ matrix is diagonalizable over $\mathbb R$ if it has a basis of two linearly independent eigenvectors. Here the eigenspace for the only eigenvalue $2$ is one-dimensional. Therefore there is only one independent eigenvector. Hence $A$ is not diagonalizable.

8.2 8.2 Nilpotent matrices: motion that eventually disappears

Definition 8.1: Nilpotent matrix

A square matrix $N$ is called nilpotent if there exists a positive integer $k$ such that

\[ N^k=0. \]

The smallest such $k$ is called the nilpotency index of $N$.

A basic nilpotent matrix is

\[ N=\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}. \]

It satisfies

\[ N^2=\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix}, \qquad N^3=0. \]

So $N$ is nilpotent of index $3$.

Interpretation

Nilpotent means “eventually zero.” A nilpotent matrix may move a vector at first, but after enough applications, every vector is sent to zero.

Proof: a nilpotent matrix has only eigenvalue $0$

Suppose $Nv=\lambda v$ with $v\neq0$. If $N^k=0$, then

\[ 0=N^kv=\lambda^k v. \]

Since $v\neq0$, this implies $\lambda^k=0$, hence $\lambda=0$.

8.3 8.3 Jordan blocks

Definition 8.2: Jordan block

For $\lambda\in\mathbb F$ and $k\ge1$, the $k\times k$ Jordan block with eigenvalue $\lambda$ is

\[ J_{\lambda,k}=\begin{bmatrix} \lambda&1&0&\cdots&0\\ 0&\lambda&1&\cdots&0\\ \vdots&\vdots&\ddots&\ddots&\vdots\\ 0&0&\cdots&\lambda&1\\ 0&0&\cdots&0&\lambda \end{bmatrix}. \]

A Jordan block can be written as

\[ J_{\lambda,k}=\lambda I+N_k, \]

where

\[ N_k=\begin{bmatrix} 0&1&0&\cdots&0\\ 0&0&1&\cdots&0\\ \vdots&\vdots&\ddots&\ddots&\vdots\\ 0&0&\cdots&0&1\\ 0&0&\cdots&0&0 \end{bmatrix}. \]

The matrix $N_k$ is nilpotent.

Theorem 8.3: Meaning of a Jordan block

On a Jordan block $J_{\lambda,k}$, the matrix action is the sum of two parts:

the diagonal eigenvalue part $\lambda I$;
the nilpotent shift part $N_k$.

Thus a Jordan block represents an eigendirection together with a chain of generalized directions.

Proof idea

The formula $J_{\lambda,k}=\lambda I+N_k$ is immediate from the entries. The diagonal entries are $\lambda$, and the only off-diagonal nonzero entries are the $1$’s just above the diagonal. Those $1$’s form the nilpotent shift $N_k$.

8.4 8.4 Generalized eigenvectors and Jordan chains

Definition 8.4: Generalized eigenvector

Let $A\in\mathbb F^{n\times n}$ and let $\lambda$ be an eigenvalue of $A$. A nonzero vector $v$ is a generalized eigenvector for $\lambda$ if

\[ (A-\lambda I)^m v=0 \]

for some positive integer $m$.

Ordinary eigenvectors correspond to $m=1$.

Definition 8.5: Jordan chain

A sequence of nonzero vectors

\[ v_1,v_2,\ldots,v_k \]

is called a Jordan chain for $A$ with eigenvalue $\lambda$ if

\[ (A-\lambda I)v_1=0, \]

and

\[ (A-\lambda I)v_j=v_{j-1},\qquad j=2,\ldots,k. \]

In this chain, $v_1$ is an ordinary eigenvector. The vectors $v_2,\ldots,v_k$ are generalized eigenvectors.

Proposition 8.6: Matrix of a Jordan chain

If $v_1,\ldots,v_k$ is a Jordan chain for $A$ with eigenvalue $\lambda$, then the matrix of $A$ in the ordered basis

\[ v_1,v_2,\ldots,v_k \]

is the Jordan block $J_{\lambda,k}$.

Proof

From the chain conditions,

\[ (A-\lambda I)v_1=0, \qquad (A-\lambda I)v_j=v_{j-1}\quad (j\ge2). \]

Therefore

\[ Av_1=\lambda v_1, \]

and

\[ Av_j=v_{j-1}+\lambda v_j\quad (j\ge2). \]

These equations say that the coordinate columns of $A$ in the basis $v_1,\ldots,v_k$ are exactly the columns of the Jordan block.

8.5 8.5 Generalized eigenspaces

Definition 8.7: Generalized eigenspace

The generalized eigenspace of $A$ associated with $\lambda$ is

\[ G_\lambda=\ker\bigl((A-\lambda I)^n\bigr), \]

where $A$ is an $n\times n$ matrix.

The ordinary eigenspace is

\[ E_\lambda=\ker(A-\lambda I). \]

The generalized eigenspace may be larger. It contains all vectors that eventually become eigenvectors, then eventually become zero after repeated application of $A-\lambda I$.

Theorem 8.8: Generalized eigenspace decomposition

Assume the characteristic polynomial of $A\in\mathbb C^{n\times n}$ splits as

\[ p_A(t)=\prod_{i=1}^s(t-\lambda_i)^{a_i}. \]

Then

\[ \mathbb C^n=G_{\lambda_1}\oplus\cdots\oplus G_{\lambda_s}. \]

Proof idea

The main idea is that different factors $(t-\lambda_i)^{a_i}$ of the characteristic polynomial are relatively prime. Using polynomial identities, one builds projection operators onto the corresponding generalized eigenspaces. This separates the whole space into independent generalized eigenspace components.

8.6 8.6 Jordan canonical form

Theorem 8.9: Jordan canonical form

Let $A\in\mathbb C^{n\times n}$. Then there exists an invertible matrix $P$ such that

\[ A=PJP^{-1}, \]

where $J$ is block diagonal and each block is a Jordan block:

\[ J=\begin{bmatrix} J_{\lambda_1,k_1}&&0\\ &\ddots&\\ 0&&J_{\lambda_r,k_r} \end{bmatrix}. \]

The matrix $J$ is called a Jordan canonical form of $A$.

The theorem says that every complex square matrix can be understood by generalized eigenvector chains.

Story

Diagonalization asks for enough eigenvectors. Jordan form asks for enough generalized eigenvectors. Over $\mathbb C$, generalized eigenvectors always provide a basis.

Proof idea

The proof has two layers. First, decompose $\mathbb C^n$ into generalized eigenspaces. Second, on each generalized eigenspace, study the nilpotent operator $N=A-\lambda I$. Nilpotent operators can be organized into chains. Each chain gives one Jordan block.

8.7 8.7 Diagonalization as a special case

A diagonal matrix is a Jordan form in which every Jordan block has size $1$:

\[ J_{\lambda,1}=[\lambda]. \]

Theorem 8.10: Diagonalizability and Jordan blocks

A matrix $A$ is diagonalizable over $\mathbb C$ if and only if every Jordan block of $A$ has size $1$.

Proof

If every Jordan block has size $1$, then the Jordan matrix is diagonal, so $A$ is similar to a diagonal matrix. Conversely, if $A$ is diagonalizable, it has a basis of ordinary eigenvectors. In such a basis, the matrix is diagonal, so no Jordan block of size larger than $1$ can appear.

8.8 8.8 How to find Jordan block sizes

Let

\[ N=A-\lambda I. \]

Define

\[ d_j=\dim\ker(N^j),\qquad j=1,2,\ldots. \]

Theorem 8.11: Kernel dimensions detect block sizes

For a fixed eigenvalue $\lambda$, the number

\[ d_j-d_{j-1} \]

equals the number of Jordan blocks for $\lambda$ of size at least $j$, where $d_0=0$.

Therefore the number of Jordan blocks of size exactly $j$ is

\[ (d_j-d_{j-1})-(d_{j+1}-d_j). \]

Proof idea

For one Jordan block of size $k$, the nilpotent part $N$ satisfies

\[ \dim\ker(N^j)=\min(j,k). \]

Thus this block contributes $1$ to $d_j-d_{j-1}$ exactly when $j\le k$. Adding over all blocks gives the formula.

8.8.1 Example 8.12: block sizes from kernel dimensions

Suppose the algebraic multiplicity of $\lambda$ is $5$ and

\[ d_1=2, \qquad d_2=4, \qquad d_3=5. \]

Then

\[ d_1-d_0=2, \qquad d_2-d_1=2, \qquad d_3-d_2=1. \]

So there are two blocks of size at least $1$, two blocks of size at least $2$, and one block of size at least $3$. Therefore the block sizes are

\[ 3 \quad\text{and}\quad 2. \]

8.9 8.9 Python computation: Jordan blocks and powers

The following code constructs a Jordan block and compares direct powers with the binomial formula.

Code

import sympy as sp

lam = sp.Symbol('lambda')
k = 3
N = sp.zeros(k)
for i in range(k-1):
    N[i, i+1] = 1
J = lam*sp.eye(k) + N
J

$\displaystyle \left[\begin{matrix}\lambda & 1 & 0\\0 & \lambda & 1\\0 & 0 & \lambda\end{matrix}\right]$

For a Jordan block,

\[ J_{\lambda,k}^m=(\lambda I+N)^m =\sum_{r=0}^{k-1}\binom{m}{r}\lambda^{m-r}N^r. \]

Code

lam_value = 2
m = 5
J2 = J.subs(lam, lam_value)
J2**m

$\displaystyle \left[\begin{matrix}32 & 80 & 80\\0 & 32 & 80\\0 & 0 & 32\end{matrix}\right]$

8.10 8.10 Powers of a Jordan block

Proposition 8.13: Powers of a Jordan block

Let

\[ J=J_{\lambda,k}=\lambda I+N, \]

where $N^k=0$. Then for every positive integer $m$,

\[ J^m=\sum_{r=0}^{k-1}\binom{m}{r}\lambda^{m-r}N^r. \]

Proof

Since $\lambda I$ commutes with $N$, the binomial theorem gives

\[ (\lambda I+N)^m=\sum_{r=0}^{m}\binom{m}{r}(\lambda I)^{m-r}N^r. \]

But $N^r=0$ for $r\ge k$. Therefore only the terms $r=0,1,\ldots,k-1$ remain.

8.10.1 Example 8.14: a $2\times2$ block

Let

\[ J=\begin{bmatrix}\lambda&1\\0&\lambda\end{bmatrix}. \]

Then

\[ J^m=\begin{bmatrix}\lambda^m&m\lambda^{m-1}\\0&\lambda^m\end{bmatrix}. \]

The off-diagonal term $m\lambda^{m-1}$ is the polynomial correction caused by the missing eigenvector.

8.11 8.11 Matrix exponentials

Matrix exponentials appear in differential equations

\[ \frac{d}{dt}x(t)=Ax(t), \qquad x(t)=e^{tA}x(0). \]

For a Jordan block,

\[ e^{tJ}=e^{t(\lambda I+N)}=e^{\lambda t}e^{tN}. \]

Since $N^k=0$,

\[ e^{tN}=I+tN+\frac{t^2}{2!}N^2+\cdots+\frac{t^{k-1}}{(k-1)!}N^{k-1}. \]

Proposition 8.15: Exponential of a Jordan block

If $J=J_{\lambda,k}=\lambda I+N$, then

\[ e^{tJ}=e^{\lambda t}\sum_{r=0}^{k-1}\frac{t^r}{r!}N^r. \]

Proof

The matrices $\lambda I$ and $N$ commute. Hence

\[ e^{t(\lambda I+N)}=e^{t\lambda I}e^{tN}=e^{\lambda t}e^{tN}. \]

Because $N$ is nilpotent, the exponential series for $e^{tN}$ stops after finitely many terms.

8.12 8.12 Minimal polynomial and largest Jordan blocks

Definition 8.16: Minimal polynomial

The minimal polynomial of a square matrix $A$ is the monic polynomial $m_A(t)$ of least degree such that

\[ m_A(A)=0. \]

Theorem 8.17: Minimal polynomial and Jordan blocks

Suppose the Jordan form of $A$ has eigenvalues $\lambda_1,\ldots,\lambda_s$. For each eigenvalue $\lambda_i$, let $r_i$ be the size of the largest Jordan block associated with $\lambda_i$. Then

\[ m_A(t)=\prod_{i=1}^s(t-\lambda_i)^{r_i}. \]

Proof idea

A Jordan block of size $r$ for eigenvalue $\lambda$ is killed by $(t-\lambda)^r$ but not by any smaller power. If several blocks share the same eigenvalue, the largest block requires the largest exponent. Multiplying over all distinct eigenvalues kills all blocks.

8.13 8.13 Numerical warning

Jordan form is powerful for exact theory, but it is unstable for numerical computation. A tiny perturbation can change the Jordan structure.

For example,

\[ \begin{bmatrix}1&1\\0&1\end{bmatrix} \]

is not diagonalizable, but

\[ \begin{bmatrix}1&1\\0&1+\varepsilon\end{bmatrix} \]

has two distinct eigenvalues when $\varepsilon\ne0$, so it is diagonalizable.

Practical message

Jordan form is a theoretical microscope. In numerical linear algebra, stable tools such as Schur decomposition, QR methods, and SVD are often preferred.

8.14 8.14 AI companion activities

8.14.1 Activity A: Explain generalized eigenvectors

Ask an AI tool:

Explain the difference between an eigenvector and a generalized eigenvector using the matrix $\begin{bmatrix}2&1\\0&2\end{bmatrix}$.

Then verify its answer by computing $\ker(A-2I)$ and $\ker((A-2I)^2)$.

8.14.2 Activity B: Find the chain

Ask:

For $A=\begin{bmatrix}2&1\\0&2\end{bmatrix}$, find a Jordan chain and explain why it gives a Jordan block.

Check that

\[ (A-2I)e_1=0, \qquad (A-2I)e_2=e_1. \]

8.14.3 Activity C: Compare diagonalization and Jordan form

Ask:

Why does a nontrivial Jordan block create polynomial factors in $A^m$ and $e^{tA}$?

Then connect the answer to the formulas in Sections 8.10 and 8.11.

8.15 8.15 Challenge questions

8.15.1 Challenge 1

Let

\[ A=\begin{bmatrix}3&1&0\\0&3&1\\0&0&3\end{bmatrix}. \]

Find $A^m$.

Solution

Write $A=3I+N$, where

\[ N=\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}, \qquad N^3=0. \]

Therefore

\[ A^m=3^mI+m3^{m-1}N+\binom{m}{2}3^{m-2}N^2. \]

Thus

\[ A^m= \begin{bmatrix} 3^m&m3^{m-1}&\binom{m}{2}3^{m-2}\\ 0&3^m&m3^{m-1}\\ 0&0&3^m \end{bmatrix}. \]

8.15.2 Challenge 2

Suppose for one eigenvalue $\lambda$ the kernel dimensions are

\[ d_1=3, \qquad d_2=5, \qquad d_3=6. \]

Find the Jordan block sizes.

Solution

We compute

\[ d_1-d_0=3, \qquad d_2-d_1=2, \qquad d_3-d_2=1. \]

So there are three blocks of size at least $1$, two blocks of size at least $2$, and one block of size at least $3$. Hence the block sizes are

\[ 3,2,1. \]

8.16 8.16 Practice problems

8.16.1 Problem 1

Determine whether

\[ A=\begin{bmatrix}4&1\\0&4\end{bmatrix} \]

is diagonalizable.

Solution

The only eigenvalue is $4$. Since

\[ A-4I=\begin{bmatrix}0&1\\0&0\end{bmatrix}, \]

the eigenspace is one-dimensional. Therefore $A$ is not diagonalizable.

8.16.2 Problem 2

Find a Jordan chain for

\[ A=\begin{bmatrix}4&1\\0&4\end{bmatrix}. \]

Solution

Let $v_1=e_1$ and $v_2=e_2$. Then

\[ (A-4I)v_1=0, \qquad (A-4I)v_2=v_1. \]

Thus $v_1,v_2$ is a Jordan chain.

8.16.3 Problem 3

Let $J=J_{2,2}$. Compute $J^5$.

Solution

For a $2\times2$ Jordan block,

\[ J^m=\begin{bmatrix}2^m&m2^{m-1}\\0&2^m\end{bmatrix}. \]

Thus

\[ J^5=\begin{bmatrix}32&80\\0&32\end{bmatrix}. \]

8.16.4 Problem 4

If the largest Jordan block for eigenvalue $1$ has size $3$ and the largest Jordan block for eigenvalue $-2$ has size $1$, find the minimal polynomial.

Solution

The minimal polynomial is

\[ m_A(t)=(t-1)^3(t+2). \]

8.17 8.17 Summary

Jordan canonical form explains the precise structure behind the failure of diagonalization.

Concept	Meaning
Eigenvector	A vector killed by $A-\lambda I$
Generalized eigenvector	A vector eventually killed by powers of $A-\lambda I$
Jordan chain	A sequence moved downward by $A-\lambda I$
Jordan block	The matrix representation of one chain
Diagonalizable matrix	All Jordan blocks have size $1$
Minimal polynomial	Records the largest Jordan block for each eigenvalue

The central message is:

\[ \boxed{\text{Jordan form replaces missing eigenvectors with generalized eigenvector chains.}} \]

--- title: "Chapter 8. When Diagonalization Fails: Jordan Canonical Form" subtitle: "Generalized eigenvectors, nilpotent motion, and the hidden structure of repeated eigenvalues" format: html: toc: true toc-depth: 3 number-sections: true code-fold: show code-tools: true html-math-method: mathjax pdf: toc: true number-sections: true jupyter: python3 --- > **Guiding question.** > If a matrix does not have enough eigenvectors to be diagonalized, what is the next best coordinate system? In Chapter 7, diagonalization told a beautiful story. If a matrix has enough eigenvectors, then we can choose an eigenvector basis and the matrix becomes diagonal: $$ A=PDP^{-1}. $$ In that coordinate system, every coordinate evolves independently. Powers, exponentials, and dynamical systems become easy. But not every matrix has enough eigenvectors. A repeated eigenvalue may give only one eigendirection. At first, this looks like failure. Jordan canonical form explains that the failure is structured. Missing eigenvectors are replaced by **generalized eigenvectors**, and these generalized eigenvectors form **chains**. Each chain becomes one **Jordan block**. ::: {.callout-note title="Main idea"} Jordan form is diagonalization plus correction terms for missing eigenvectors: $$ \boxed{A=PJP^{-1},\qquad J=\text{block diagonal matrix of Jordan blocks}.} $$ A diagonal matrix is the special case in which every Jordan block has size $1$. ::: ::: {.callout-tip title="AI and coding companion"} Ask an AI tool to explain the difference between an eigenvector and a generalized eigenvector using a simple $2\times2$ Jordan block. Then verify the formulas by hand and with Python. ::: ## 8.1 The obstruction: not enough eigenvectors Consider $$ A=\begin{bmatrix}2&1\\0&2\end{bmatrix}. $$ The only eigenvalue is $\lambda=2$ because $$ \det(A-\lambda I)=(2-\lambda)^2. $$ But $$ A-2I=\begin{bmatrix}0&1\\0&0\end{bmatrix}, $$ so $$ \ker(A-2I)=\left\{\begin{bmatrix}x\\0\end{bmatrix}:x\in\mathbb R\right\}. $$ Thus the eigenspace is one-dimensional, even though the algebraic multiplicity of $2$ is two. We do not have enough eigenvectors to diagonalize $A$. ::: {.callout-warning title="Warning"} A repeated eigenvalue does not automatically cause a problem. The problem occurs when the eigenspace is too small. ::: <details> <summary>Proof: why this matrix is not diagonalizable</summary> A $2\times2$ matrix is diagonalizable over $\mathbb R$ if it has a basis of two linearly independent eigenvectors. Here the eigenspace for the only eigenvalue $2$ is one-dimensional. Therefore there is only one independent eigenvector. Hence $A$ is not diagonalizable. </details> ## 8.2 Nilpotent matrices: motion that eventually disappears ::: {.callout-note title="Definition 8.1: Nilpotent matrix"} A square matrix $N$ is called **nilpotent** if there exists a positive integer $k$ such that $$ N^k=0. $$ The smallest such $k$ is called the **nilpotency index** of $N$. ::: A basic nilpotent matrix is $$ N=\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}. $$ It satisfies $$ N^2=\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix}, \qquad N^3=0. $$ So $N$ is nilpotent of index $3$. ::: {.callout-important title="Interpretation"} Nilpotent means “eventually zero.” A nilpotent matrix may move a vector at first, but after enough applications, every vector is sent to zero. ::: <details> <summary>Proof: a nilpotent matrix has only eigenvalue $0$</summary> Suppose $Nv=\lambda v$ with $v\neq0$. If $N^k=0$, then $$ 0=N^kv=\lambda^k v. $$ Since $v\neq0$, this implies $\lambda^k=0$, hence $\lambda=0$. </details> ## 8.3 Jordan blocks ::: {.callout-note title="Definition 8.2: Jordan block"} For $\lambda\in\mathbb F$ and $k\ge1$, the $k\times k$ **Jordan block** with eigenvalue $\lambda$ is $$ J_{\lambda,k}=\begin{bmatrix} \lambda&1&0&\cdots&0\\ 0&\lambda&1&\cdots&0\\ \vdots&\vdots&\ddots&\ddots&\vdots\\ 0&0&\cdots&\lambda&1\\ 0&0&\cdots&0&\lambda \end{bmatrix}. $$ ::: A Jordan block can be written as $$ J_{\lambda,k}=\lambda I+N_k, $$ where $$ N_k=\begin{bmatrix} 0&1&0&\cdots&0\\ 0&0&1&\cdots&0\\ \vdots&\vdots&\ddots&\ddots&\vdots\\ 0&0&\cdots&0&1\\ 0&0&\cdots&0&0 \end{bmatrix}. $$ The matrix $N_k$ is nilpotent. ::: {.callout-note title="Theorem 8.3: Meaning of a Jordan block"} On a Jordan block $J_{\lambda,k}$, the matrix action is the sum of two parts: 1. the diagonal eigenvalue part $\lambda I$; 2. the nilpotent shift part $N_k$. Thus a Jordan block represents an eigendirection together with a chain of generalized directions. ::: <details> <summary>Proof idea</summary> The formula $J_{\lambda,k}=\lambda I+N_k$ is immediate from the entries. The diagonal entries are $\lambda$, and the only off-diagonal nonzero entries are the $1$'s just above the diagonal. Those $1$'s form the nilpotent shift $N_k$. </details> ## 8.4 Generalized eigenvectors and Jordan chains ::: {.callout-note title="Definition 8.4: Generalized eigenvector"} Let $A\in\mathbb F^{n\times n}$ and let $\lambda$ be an eigenvalue of $A$. A nonzero vector $v$ is a **generalized eigenvector** for $\lambda$ if $$ (A-\lambda I)^m v=0 $$ for some positive integer $m$. ::: Ordinary eigenvectors correspond to $m=1$. ::: {.callout-note title="Definition 8.5: Jordan chain"} A sequence of nonzero vectors $$ v_1,v_2,\ldots,v_k $$ is called a **Jordan chain** for $A$ with eigenvalue $\lambda$ if $$ (A-\lambda I)v_1=0, $$ and $$ (A-\lambda I)v_j=v_{j-1},\qquad j=2,\ldots,k. $$ ::: In this chain, $v_1$ is an ordinary eigenvector. The vectors $v_2,\ldots,v_k$ are generalized eigenvectors. ::: {.callout-note title="Proposition 8.6: Matrix of a Jordan chain"} If $v_1,\ldots,v_k$ is a Jordan chain for $A$ with eigenvalue $\lambda$, then the matrix of $A$ in the ordered basis $$ v_1,v_2,\ldots,v_k $$ is the Jordan block $J_{\lambda,k}$. ::: <details> <summary>Proof</summary> From the chain conditions, $$ (A-\lambda I)v_1=0, \qquad (A-\lambda I)v_j=v_{j-1}\quad (j\ge2). $$ Therefore $$ Av_1=\lambda v_1, $$ and $$ Av_j=v_{j-1}+\lambda v_j\quad (j\ge2). $$ These equations say that the coordinate columns of $A$ in the basis $v_1,\ldots,v_k$ are exactly the columns of the Jordan block. </details> ## 8.5 Generalized eigenspaces ::: {.callout-note title="Definition 8.7: Generalized eigenspace"} The **generalized eigenspace** of $A$ associated with $\lambda$ is $$ G_\lambda=\ker\bigl((A-\lambda I)^n\bigr), $$ where $A$ is an $n\times n$ matrix. ::: The ordinary eigenspace is $$ E_\lambda=\ker(A-\lambda I). $$ The generalized eigenspace may be larger. It contains all vectors that eventually become eigenvectors, then eventually become zero after repeated application of $A-\lambda I$. ::: {.callout-note title="Theorem 8.8: Generalized eigenspace decomposition"} Assume the characteristic polynomial of $A\in\mathbb C^{n\times n}$ splits as $$ p_A(t)=\prod_{i=1}^s(t-\lambda_i)^{a_i}. $$ Then $$ \mathbb C^n=G_{\lambda_1}\oplus\cdots\oplus G_{\lambda_s}. $$ ::: <details> <summary>Proof idea</summary> The main idea is that different factors $(t-\lambda_i)^{a_i}$ of the characteristic polynomial are relatively prime. Using polynomial identities, one builds projection operators onto the corresponding generalized eigenspaces. This separates the whole space into independent generalized eigenspace components. </details> ## 8.6 Jordan canonical form ::: {.callout-note title="Theorem 8.9: Jordan canonical form"} Let $A\in\mathbb C^{n\times n}$. Then there exists an invertible matrix $P$ such that $$ A=PJP^{-1}, $$ where $J$ is block diagonal and each block is a Jordan block: $$ J=\begin{bmatrix} J_{\lambda_1,k_1}&&0\\ &\ddots&\\ 0&&J_{\lambda_r,k_r} \end{bmatrix}. $$ The matrix $J$ is called a **Jordan canonical form** of $A$. ::: The theorem says that every complex square matrix can be understood by generalized eigenvector chains. ::: {.callout-important title="Story"} Diagonalization asks for enough eigenvectors. Jordan form asks for enough generalized eigenvectors. Over $\mathbb C$, generalized eigenvectors always provide a basis. ::: <details> <summary>Proof idea</summary> The proof has two layers. First, decompose $\mathbb C^n$ into generalized eigenspaces. Second, on each generalized eigenspace, study the nilpotent operator $N=A-\lambda I$. Nilpotent operators can be organized into chains. Each chain gives one Jordan block. </details> ## 8.7 Diagonalization as a special case A diagonal matrix is a Jordan form in which every Jordan block has size $1$: $$ J_{\lambda,1}=[\lambda]. $$ ::: {.callout-note title="Theorem 8.10: Diagonalizability and Jordan blocks"} A matrix $A$ is diagonalizable over $\mathbb C$ if and only if every Jordan block of $A$ has size $1$. ::: <details> <summary>Proof</summary> If every Jordan block has size $1$, then the Jordan matrix is diagonal, so $A$ is similar to a diagonal matrix. Conversely, if $A$ is diagonalizable, it has a basis of ordinary eigenvectors. In such a basis, the matrix is diagonal, so no Jordan block of size larger than $1$ can appear. </details> ## 8.8 How to find Jordan block sizes Let $$ N=A-\lambda I. $$ Define $$ d_j=\dim\ker(N^j),\qquad j=1,2,\ldots. $$ ::: {.callout-note title="Theorem 8.11: Kernel dimensions detect block sizes"} For a fixed eigenvalue $\lambda$, the number $$ d_j-d_{j-1} $$ equals the number of Jordan blocks for $\lambda$ of size at least $j$, where $d_0=0$. Therefore the number of Jordan blocks of size exactly $j$ is $$ (d_j-d_{j-1})-(d_{j+1}-d_j). $$ ::: <details> <summary>Proof idea</summary> For one Jordan block of size $k$, the nilpotent part $N$ satisfies $$ \dim\ker(N^j)=\min(j,k). $$ Thus this block contributes $1$ to $d_j-d_{j-1}$ exactly when $j\le k$. Adding over all blocks gives the formula. </details> ### Example 8.12: block sizes from kernel dimensions Suppose the algebraic multiplicity of $\lambda$ is $5$ and $$ d_1=2, \qquad d_2=4, \qquad d_3=5. $$ Then $$ d_1-d_0=2, \qquad d_2-d_1=2, \qquad d_3-d_2=1. $$ So there are two blocks of size at least $1$, two blocks of size at least $2$, and one block of size at least $3$. Therefore the block sizes are $$ 3 \quad\text{and}\quad 2. $$ ## 8.9 Python computation: Jordan blocks and powers The following code constructs a Jordan block and compares direct powers with the binomial formula. ```{python} import sympy as sp lam = sp.Symbol('lambda') k = 3 N = sp.zeros(k) for i in range(k-1): N[i, i+1] = 1 J = lam*sp.eye(k) + N J ``` For a Jordan block, $$ J_{\lambda,k}^m=(\lambda I+N)^m =\sum_{r=0}^{k-1}\binom{m}{r}\lambda^{m-r}N^r. $$ ```{python} lam_value = 2 m = 5 J2 = J.subs(lam, lam_value) J2**m ``` ## 8.10 Powers of a Jordan block ::: {.callout-note title="Proposition 8.13: Powers of a Jordan block"} Let $$ J=J_{\lambda,k}=\lambda I+N, $$ where $N^k=0$. Then for every positive integer $m$, $$ J^m=\sum_{r=0}^{k-1}\binom{m}{r}\lambda^{m-r}N^r. $$ ::: <details> <summary>Proof</summary> Since $\lambda I$ commutes with $N$, the binomial theorem gives $$ (\lambda I+N)^m=\sum_{r=0}^{m}\binom{m}{r}(\lambda I)^{m-r}N^r. $$ But $N^r=0$ for $r\ge k$. Therefore only the terms $r=0,1,\ldots,k-1$ remain. </details> ### Example 8.14: a $2\times2$ block Let $$ J=\begin{bmatrix}\lambda&1\\0&\lambda\end{bmatrix}. $$ Then $$ J^m=\begin{bmatrix}\lambda^m&m\lambda^{m-1}\\0&\lambda^m\end{bmatrix}. $$ The off-diagonal term $m\lambda^{m-1}$ is the polynomial correction caused by the missing eigenvector. ## 8.11 Matrix exponentials Matrix exponentials appear in differential equations $$ \frac{d}{dt}x(t)=Ax(t), \qquad x(t)=e^{tA}x(0). $$ For a Jordan block, $$ e^{tJ}=e^{t(\lambda I+N)}=e^{\lambda t}e^{tN}. $$ Since $N^k=0$, $$ e^{tN}=I+tN+\frac{t^2}{2!}N^2+\cdots+\frac{t^{k-1}}{(k-1)!}N^{k-1}. $$ ::: {.callout-note title="Proposition 8.15: Exponential of a Jordan block"} If $J=J_{\lambda,k}=\lambda I+N$, then $$ e^{tJ}=e^{\lambda t}\sum_{r=0}^{k-1}\frac{t^r}{r!}N^r. $$ ::: <details> <summary>Proof</summary> The matrices $\lambda I$ and $N$ commute. Hence $$ e^{t(\lambda I+N)}=e^{t\lambda I}e^{tN}=e^{\lambda t}e^{tN}. $$ Because $N$ is nilpotent, the exponential series for $e^{tN}$ stops after finitely many terms. </details> ## 8.12 Minimal polynomial and largest Jordan blocks ::: {.callout-note title="Definition 8.16: Minimal polynomial"} The **minimal polynomial** of a square matrix $A$ is the monic polynomial $m_A(t)$ of least degree such that $$ m_A(A)=0. $$ ::: ::: {.callout-note title="Theorem 8.17: Minimal polynomial and Jordan blocks"} Suppose the Jordan form of $A$ has eigenvalues $\lambda_1,\ldots,\lambda_s$. For each eigenvalue $\lambda_i$, let $r_i$ be the size of the largest Jordan block associated with $\lambda_i$. Then $$ m_A(t)=\prod_{i=1}^s(t-\lambda_i)^{r_i}. $$ ::: <details> <summary>Proof idea</summary> A Jordan block of size $r$ for eigenvalue $\lambda$ is killed by $(t-\lambda)^r$ but not by any smaller power. If several blocks share the same eigenvalue, the largest block requires the largest exponent. Multiplying over all distinct eigenvalues kills all blocks. </details> ## 8.13 Numerical warning Jordan form is powerful for exact theory, but it is unstable for numerical computation. A tiny perturbation can change the Jordan structure. For example, $$ \begin{bmatrix}1&1\\0&1\end{bmatrix} $$ is not diagonalizable, but $$ \begin{bmatrix}1&1\\0&1+\varepsilon\end{bmatrix} $$ has two distinct eigenvalues when $\varepsilon\ne0$, so it is diagonalizable. ::: {.callout-warning title="Practical message"} Jordan form is a theoretical microscope. In numerical linear algebra, stable tools such as Schur decomposition, QR methods, and SVD are often preferred. ::: ## 8.14 AI companion activities ### Activity A: Explain generalized eigenvectors Ask an AI tool: > Explain the difference between an eigenvector and a generalized eigenvector using the matrix $\begin{bmatrix}2&1\\0&2\end{bmatrix}$. Then verify its answer by computing $\ker(A-2I)$ and $\ker((A-2I)^2)$. ### Activity B: Find the chain Ask: > For $A=\begin{bmatrix}2&1\\0&2\end{bmatrix}$, find a Jordan chain and explain why it gives a Jordan block. Check that $$ (A-2I)e_1=0, \qquad (A-2I)e_2=e_1. $$ ### Activity C: Compare diagonalization and Jordan form Ask: > Why does a nontrivial Jordan block create polynomial factors in $A^m$ and $e^{tA}$? Then connect the answer to the formulas in Sections 8.10 and 8.11. ## 8.15 Challenge questions ### Challenge 1 Let $$ A=\begin{bmatrix}3&1&0\\0&3&1\\0&0&3\end{bmatrix}. $$ Find $A^m$. <details> <summary>Solution</summary> Write $A=3I+N$, where $$ N=\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}, \qquad N^3=0. $$ Therefore $$ A^m=3^mI+m3^{m-1}N+\binom{m}{2}3^{m-2}N^2. $$ Thus $$ A^m= \begin{bmatrix} 3^m&m3^{m-1}&\binom{m}{2}3^{m-2}\\ 0&3^m&m3^{m-1}\\ 0&0&3^m \end{bmatrix}. $$ </details> ### Challenge 2 Suppose for one eigenvalue $\lambda$ the kernel dimensions are $$ d_1=3, \qquad d_2=5, \qquad d_3=6. $$ Find the Jordan block sizes. <details> <summary>Solution</summary> We compute $$ d_1-d_0=3, \qquad d_2-d_1=2, \qquad d_3-d_2=1. $$ So there are three blocks of size at least $1$, two blocks of size at least $2$, and one block of size at least $3$. Hence the block sizes are $$ 3,2,1. $$ </details> ## 8.16 Practice problems ### Problem 1 Determine whether $$ A=\begin{bmatrix}4&1\\0&4\end{bmatrix} $$ is diagonalizable. <details> <summary>Solution</summary> The only eigenvalue is $4$. Since $$ A-4I=\begin{bmatrix}0&1\\0&0\end{bmatrix}, $$ the eigenspace is one-dimensional. Therefore $A$ is not diagonalizable. </details> ### Problem 2 Find a Jordan chain for $$ A=\begin{bmatrix}4&1\\0&4\end{bmatrix}. $$ <details> <summary>Solution</summary> Let $v_1=e_1$ and $v_2=e_2$. Then $$ (A-4I)v_1=0, \qquad (A-4I)v_2=v_1. $$ Thus $v_1,v_2$ is a Jordan chain. </details> ### Problem 3 Let $J=J_{2,2}$. Compute $J^5$. <details> <summary>Solution</summary> For a $2\times2$ Jordan block, $$ J^m=\begin{bmatrix}2^m&m2^{m-1}\\0&2^m\end{bmatrix}. $$ Thus $$ J^5=\begin{bmatrix}32&80\\0&32\end{bmatrix}. $$ </details> ### Problem 4 If the largest Jordan block for eigenvalue $1$ has size $3$ and the largest Jordan block for eigenvalue $-2$ has size $1$, find the minimal polynomial. <details> <summary>Solution</summary> The minimal polynomial is $$ m_A(t)=(t-1)^3(t+2). $$ </details> ## 8.17 Summary Jordan canonical form explains the precise structure behind the failure of diagonalization. | Concept | Meaning | |---|---| | Eigenvector | A vector killed by $A-\lambda I$ | | Generalized eigenvector | A vector eventually killed by powers of $A-\lambda I$ | | Jordan chain | A sequence moved downward by $A-\lambda I$ | | Jordan block | The matrix representation of one chain | | Diagonalizable matrix | All Jordan blocks have size $1$ | | Minimal polynomial | Records the largest Jordan block for each eigenvalue | The central message is: $$ \boxed{\text{Jordan form replaces missing eigenvectors with generalized eigenvector chains.}} $$

8.1 8.1 The obstruction: not enough eigenvectors

8.2 8.2 Nilpotent matrices: motion that eventually disappears

8.3 8.3 Jordan blocks

8.4 8.4 Generalized eigenvectors and Jordan chains

8.5 8.5 Generalized eigenspaces

8.6 8.6 Jordan canonical form

8.7 8.7 Diagonalization as a special case

8.8 8.8 How to find Jordan block sizes

8.8.1 Example 8.12: block sizes from kernel dimensions

8.9 8.9 Python computation: Jordan blocks and powers

8.10 8.10 Powers of a Jordan block

8.10.1 Example 8.14: a \(2\times2\) block

8.11 8.11 Matrix exponentials

8.12 8.12 Minimal polynomial and largest Jordan blocks

8.13 8.13 Numerical warning

8.14 8.14 AI companion activities

8.14.1 Activity A: Explain generalized eigenvectors

8.14.2 Activity B: Find the chain

8.14.3 Activity C: Compare diagonalization and Jordan form

8.15 8.15 Challenge questions

8.15.1 Challenge 1

8.15.2 Challenge 2

8.16 8.16 Practice problems

8.16.1 Problem 1

8.16.2 Problem 2

8.16.3 Problem 3

8.16.4 Problem 4

8.17 8.17 Summary