Constructing Jordan Form

Definition For $\lambda \in \mathbb{F}$ and an integer $k \ge 1$, the Jordan block $J_k(\lambda)$ is the $k \times k$ matrix with $\lambda$ in every diagonal position, $1$ in every position just above the diagonal, and $0$ elsewhere,

In particular $J_1(\lambda) = [\lambda]$. A linear operator $T$ has the matrix $J_k(\lambda)$ in an ordered basis $(v_1, \dots, v_k)$ exactly when

so the basis is a Jordan chain of length $k$ listed bottom first, from the eigenvector $v_1$ up to the top vector $v_k$. Listing the chain top first would put the ones below the diagonal instead. The matrix $J_k(\lambda) - \lambda I = J_k(0)$ is nilpotent with $J_k(0)^{k-1} \ne 0$ and $J_k(0)^k = 0$, so the characteristic and minimal polynomials of $J_k(\lambda)$ are both $(t - \lambda)^k$, and its eigenspace for $\lambda$ is the line spanned by $e_1$.

Definition Let $N \colon V \to V$ be nilpotent on a finite-dimensional space $V \ne 0$ over $\mathbb{F}$, and let $s$ be its index, the smallest $s \ge 1$ with $N^s = 0$. For $k \ge 1$ put

with $N^0 = I$. A nonzero vector of $E_k$ is killed by $N$ and equals $N^{k-1}w$ for some $w$, so it can be the bottom of a chain of length $k$. Each image $\operatorname{im} N^{k}$ lies inside $\operatorname{im} N^{k-1}$, because $N^k x = N^{k-1}(Nx)$, so the spaces are nested,

Here $E_s = \operatorname{im} N^{s-1}$ is nonzero because $N^{s-1} \ne 0$, and it lies in $\ker N$ because $N^s = 0$. The top-down construction chooses a basis of $E_s$, extends it to a basis of $E_{s-1}$, and continues down to a basis of $E_1$, always keeping the vectors already chosen. A chosen vector has entry level $k$ if it first appears in the chosen basis of $E_k$. Two facts follow at once. For every $h \ge 1$ the chosen vectors with entry level at least $h$ form a basis of $E_h$, and exactly $\dim E_k - \dim E_{k+1}$ chosen vectors have entry level $k$.

Definition Let $b$ be a vector of the top-down construction with entry level $k$. Since $b \in \operatorname{im} N^{k-1}$, there is a vector $w \in V$ with $N^{k-1}w = b$, found by solving one linear system. Lifting $b$ means choosing such a top $w$ and forming the chain

listed bottom first. Its bottom is $b$. Because $b \ne 0$ and $Nb = 0$, the top satisfies $N^{k}w = 0$ and $N^{k-1}w \ne 0$, so $w$ is a generalized eigenvector of rank exactly $k$ for the eigenvalue $0$, and none of the $k$ chain vectors is zero. For $k = 1$ the chain is just $(b)$, and its top $w = b$ is forced. For $k \ge 2$ the top is not unique, since $w + z$ works equally well for every $z$ in $\ker N^{k-1}$, which contains the nonzero space $\ker N$.

Definition Let $T \colon V \to V$ be linear on a finite-dimensional space over $\mathbb{F}$. A Jordan chain of $T$ for $\lambda \in \mathbb{F}$ is a list $(v_1, \dots, v_k)$ with $v_1 \ne 0$, $(T - \lambda I)v_1 = 0$ and $(T - \lambda I)v_j = v_{j-1}$ for $2 \le j \le k$. A Jordan basis of $T$ is an ordered basis of $V$ made by listing the vectors of finitely many Jordan chains of $T$, one chain after another, each bottom first. The matrix of $T$ in a Jordan basis is block diagonal, with one block $J_k(\lambda)$ for each chain of length $k$ for $\lambda$. For a nilpotent operator every chain has $\lambda = 0$.

The construction rests on one theorem. Let $N$ be nilpotent on a finite-dimensional space $V \ne 0$ over $\mathbb{F}$, and lift every vector of the top-down construction, a vector of entry level $k$ to a chain of length $k$, with any choice of tops. Then all the vectors of all these chains together are linearly independent. To prove it, call the $j$th vector of a chain, counted from the bottom, its level. Suppose a linear combination of all the chain vectors is $0$ with some coefficient nonzero, and let $h$ be the highest level carrying a nonzero coefficient. Apply $N^{h-1}$. A vector at a level below $h$ goes to $0$, a vector at level $h$ goes to the bottom of its chain, and every term above level $h$ already has coefficient $0$. What remains is a vanishing combination of the bottoms of the chains of length at least $h$, with the level $h$ coefficients. Those bottoms are the vectors of entry level at least $h$, which form a basis of $E_h$, so all these coefficients are zero. That contradicts the choice of $h$, so every coefficient is zero.

Definition Let $N$ be nilpotent of index $s$ on $V$ with $\dim V = n \ge 1$, and let $d_k = \dim E_k$, so that $d_{s+1} = 0$. The construction gives $d_k - d_{k+1}$ chains of length exactly $k$. Counting their vectors chain by chain, or level by level with $d_k$ vectors at level $k$, gives the same total,

To find $d_k$ without knowing any chains, restrict $N$ to $\operatorname{im} N^{k-1}$. This restriction maps onto $\operatorname{im} N^k$ and its kernel is $E_k$, so rank-nullity gives

The sum telescopes,

So the chain vectors form an independent list of $n = \dim V$ vectors, which is a basis of $V$. Every nilpotent operator on a nonzero finite-dimensional space over $\mathbb{R}$ or $\mathbb{C}$ therefore has a Jordan basis, and this nilpotent case needs no splitting assumption.

Definition A Jordan matrix is a block diagonal matrix whose diagonal blocks are Jordan blocks. A Jordan form of a linear operator $T$ is a Jordan matrix that represents $T$ in some basis, and a Jordan form of a square matrix $A$ is a Jordan matrix similar to $A$ over $\mathbb{F}$. The existence theorem reads as follows. Let $V \ne 0$ be finite-dimensional over $\mathbb{F}$ and let $T \colon V \to V$ be linear with minimal polynomial

where $\lambda_1, \dots, \lambda_r \in \mathbb{F}$ are distinct and every $s_j \ge 1$, that is, $\mu_T$ splits over $\mathbb{F}$. Then $V$ has a Jordan basis $\mathcal{B}$ for $T$, so $[T]_{\mathcal{B} \leftarrow \mathcal{B}}$ is a Jordan matrix. For a matrix $A \in \mathbb{F}^{n \times n}$ whose minimal polynomial splits, this says $P^{-1}AP = J$ for a Jordan matrix $J$ and an invertible $P \in \mathbb{F}^{n \times n}$ whose columns form a Jordan basis. Over $\mathbb{C}$ the hypothesis always holds, by the fundamental theorem of algebra. The hypothesis is also necessary. A Jordan matrix is upper triangular, so if $T$ has a Jordan basis, then $\chi_T$ is a product of linear factors over $\mathbb{F}$, and so is its divisor $\mu_T$.

The proof joins the pieces. The generalized eigenspace decomposition gives $V = G_{\lambda_1} \oplus \cdots \oplus G_{\lambda_r}$ with each $G_{\lambda_j}$ invariant under $T$, and on $G_{\lambda_j}$ the operator $N_j = T - \lambda_j I$ is nilpotent of index $s_j$. A Jordan basis of $G_{\lambda_j}$ for $N_j$ is a Jordan basis for the restriction of $T$, with eigenvalue $\lambda_j$, because $T = \lambda_j I + N_j$ there. Joining these bases gives a basis of $V$ because the sum is direct. The blocks for $\lambda_j$ fill $G_{\lambda_j}$, so their sizes add up to the algebraic multiplicity of $\lambda_j$, and the largest has size $s_j$. The Jordan basis is not unique, since every chain can be rescaled by a scalar $c \notin \{0, 1\}$. In addition, a chain of length at least $2$ can be lifted from a different top, and two or more blocks can be listed in another order.