Every student of linear algebra encounters the dream of diagonalisation: find a basis in which a linear operator acts by simply scaling each coordinate. When that dream is realised, the entire theory becomes transparent — powers of the matrix, exponentials, invariant subspaces all read off at a glance. But the dream fails more often than it succeeds. The rotation matrix

\begin{align*} R &= \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \end{align*}

over $\mathbb{R}$ has characteristic polynomial $\lambda^2 + 1$, which has no real roots, so $R$ admits no real eigenvalues at all. Even over $\mathbb{C}$, the matrix

\begin{align*} N &= \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \end{align*}

has $\lambda = 0$ as a double eigenvalue but only a one-dimensional eigenspace: $\ker N = \operatorname{span}\{e_1\}$. Any attempt to diagonalise $N$ fails because we simply do not have enough linearly independent eigenvectors to form a basis.

The Jordan Normal Form theorem says that over an algebraically closed field — most importantly $\mathbb{C}$ — every linear operator on a finite-dimensional vector space does have a canonical form, just not necessarily a diagonal one. Instead of diagonal matrices, the canonical form consists of **Jordan blocks**: triangular matrices with a single eigenvalue on the diagonal and ones on the superdiagonal. These blocks are as close to diagonal as the structure of the operator allows, and they encode every invariant of the operator up to similarity.

[example: Why Diagonalisation Fails for a Nilpotent Matrix] Let $V = \mathbb{C}^3$ and let $T: V \to V$ be given by \begin{align*} T &= \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}. \end{align*} The characteristic polynomial is $\lambda^3$, so $\lambda = 0$ is the only eigenvalue with algebraic multiplicity $3$. The eigenspace is $\ker T = \{v : Tv = 0\}$. A vector $v = (a, b, c)^\top$ lies in $\ker T$ only when $Tv = (b, c, 0)^\top = 0$, which forces $b = c = 0$. Thus $\ker T = \operatorname{span}\{e_1\}$ has dimension $1$. The geometric multiplicity (dimension of the eigenspace) is $1$, but the algebraic multiplicity is $3$. There are not enough eigenvectors to span $V$, so $T$ is not diagonalisable. Notice further that $T^2$ sends $e_3 \mapsto e_1$ and $T^3 = 0$, so the chain $e_3, T e_3 = e_2, T^2 e_3 = e_1$ forms a natural basis in which $T$ acts by shifting: $T e_3 = e_2$, $T e_2 = e_1$, $T e_1 = 0$. In this basis, $T$ is exactly the $3 \times 3$ Jordan block $J_3(0)$ — this is the Jordan Normal Form of $T$. [/example]

This example reveals the key idea: when eigenvectors are scarce, we extend them to chains. The Jordan Normal Form theorem is the precise statement that such chains always exist and together span the entire space.

[example: What Goes Wrong When You Assume Diagonalisability] Suppose one naively assumes that every matrix over $\mathbb{C}$ is diagonalisable and tries to compute $e^{tA}$ for \begin{align*} A &= \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}. \end{align*} The characteristic polynomial is $(\lambda - 1)^2$, so $\lambda = 1$ is a double eigenvalue. The eigenspace is $\ker(A - I) = \ker \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \operatorname{span}\{e_1\}$, which is one-dimensional. There is no basis of eigenvectors, so $A$ is not diagonalisable. If one ignores this and writes $e^{tA} = e^t I$ (incorrectly treating $A$ as the scalar matrix $1 \cdot I$), the "solution" $x(t) = e^t x_0$ to $\dot{x} = Ax$ fails: differentiating gives $\dot{x}(t) = e^t x_0$, but $A x(t) = e^t A x_0$, and $e^t x_0 = e^t A x_0$ would require $x_0 = A x_0$ for all initial data — which is false since $A \neq I$. The correct formula, obtained from the Jordan Normal Form $A = J_2(1)$, gives $e^{tA} = e^t \begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix}$. For $x_0 = (0, 1)^\top$: $x(t) = e^t (t, 1)^\top$. The first component $x_1(t) = t e^t$ has linear growth in $t$ multiplying the exponential — a signature of a non-trivial Jordan block that diagonalisation completely misses. [/example]

## Definition and Basic Structure

Before stating the Jordan Normal Form theorem, we need to name its building blocks.

When a linear operator $T$ has an eigenvalue $\lambda$, not all behaviour at $\lambda$ is captured by the eigenspace $\ker(T - \lambda I)$. There may be vectors that are not eigenvectors but become eigenvectors after applying $T - \lambda I$ once, or twice, or finitely many times. This "generalised" structure is what the Jordan block encodes.

[definition: Jordan Block] Let $\lambda \in \mathbb{C}$ and $m \in \mathbb{N}$. The **Jordan block** of size $m$ with eigenvalue $\lambda$ is the $m \times m$ matrix \begin{align*} J_m(\lambda) &= \begin{pmatrix} \lambda & 1 & 0 & \cdots & 0 \\ 0 & \lambda & 1 & \cdots & 0 \\ 0 & 0 & \lambda & \cdots & 0 \\ \vdots & & & \ddots & 1 \\ 0 & 0 & 0 & \cdots & \lambda \end{pmatrix} \in \mathrm{Mat}_m(\mathbb{C}). \end{align*} That is, $(J_m(\lambda))_{ij} = \lambda$ when $i = j$, $(J_m(\lambda))_{ij} = 1$ when $j = i + 1$, and $0$ otherwise. [/definition]

The Jordan block $J_m(\lambda)$ can be written as $\lambda I_m + N_m$, where $N_m$ is the $m \times m$ nilpotent shift matrix with $1$s on the superdiagonal. This decomposition is fundamental: $\lambda I$ is the scalar part, and $N_m$ is the nilpotent "error" that prevents $J_m(\lambda)$ from being a scalar matrix when $m > 1$.

[remark: Size-One Jordan Blocks] A Jordan block of size $1$ is simply $J_1(\lambda) = (\lambda)$, a $1 \times 1$ scalar matrix. An operator is diagonalisable if and only if its Jordan Normal Form consists entirely of $1 \times 1$ blocks — each eigenvalue appears with a block for each eigenvector in a basis. [/remark]

[definition: Jordan Normal Form] Let $V$ be a finite-dimensional vector space over $\mathbb{C}$ and let $T: V \to V$ be a linear operator. A **Jordan Normal Form** of $T$ is a block-diagonal matrix \begin{align*} J &= \begin{pmatrix} J_{m_1}(\lambda_1) & & \\ & \ddots & \\ & & J_{m_k}(\lambda_k) \end{pmatrix} \end{align*} where $\lambda_1, \ldots, \lambda_k \in \mathbb{C}$ (not necessarily distinct) and $m_1 + \cdots + m_k = \dim V$, such that $T$ is similar to $J$: there exists an invertible $P$ with $P^{-1}TP = J$. [/definition]

The entries $\lambda_i$ are the eigenvalues of $T$ (with repetition), and the sizes $m_i$ are the sizes of the corresponding Jordan blocks. The Jordan Normal Form is unique up to permutation of the blocks — this is the content of the uniqueness assertion in the main theorem.

[quotetheorem:412]

The proof of this theorem — which we will build toward through generalised eigenspaces and cyclic decomposition — is the central goal of this chapter. We will prove it by understanding the structure of nilpotent operators first, then combining that with the primary decomposition theorem.

## Generalised Eigenspaces and the Primary Decomposition