Computing Jordan Form and Its Limitations

Definition Let $T \colon V \to V$ be linear with $\dim V = n$, and let $\lambda \in \mathbb{F}$ be an eigenvalue of $T$. The successive kernel dimensions of $T$ at $\lambda$ are

so $d_0 = 0$, and for a matrix $A \in \mathbb{F}^{n \times n}$ they are $d_k = n - \operatorname{rank}(A - \lambda I)^k$. They strictly increase up to an index $s$ and stay constant from then on,

where $G_\lambda$ is the generalized eigenspace of $\lambda$, whose dimension is the algebraic multiplicity of $\lambda$, and $s$ is the exponent of $t - \lambda$ in the minimal polynomial. All these kernels lie inside $G_\lambda$, because $T - \lambda I$ is invertible on the generalized eigenspaces of the other eigenvalues, so the computation may use the whole space. In a Jordan basis of $T$, a block $J_\ell(\lambda)$ contributes $\min(k, \ell)$ to $d_k$, since the kernel of $J_\ell(0)^k$ is spanned by the first $\min(k, \ell)$ vectors of its chain. A block $J_\ell(\mu)$ with $\mu \ne \lambda$ contributes $0$, since $J_\ell(\mu) - \lambda I$ is triangular with nonzero diagonal entries $\mu - \lambda$. Hence

Definition With $d_k$ as above, let $c_k = d_k - d_{k-1}$ for $k \ge 1$. Then $c_k$ is the number of Jordan blocks for $\lambda$ of size at least $k$, because $\min(k, \ell) - \min(k - 1, \ell)$ is $1$ when $\ell \ge k$ and $0$ otherwise. With $N = T - \lambda I$, the increment also equals $\operatorname{rank} N^{k-1} - \operatorname{rank} N^{k}$, which is $\dim(\ker N \cap \operatorname{im} N^{k-1})$ by rank-nullity for $N$ restricted to $\operatorname{im} N^{k-1}$. So the spaces $E_k = \ker N \cap \operatorname{im} N^{k-1}$ of the top-down construction have dimensions $c_k$, not $d_k$. The number of blocks for $\lambda$ of size exactly $k$ is therefore

Several checks come with this rule. The increments satisfy $c_1 \ge c_2 \ge \cdots \ge 0$. The number of blocks is $c_1 = d_1 = \dim \ker(T - \lambda I)$, the geometric multiplicity of $\lambda$. The block sizes add up to $d_s$, the algebraic multiplicity. The largest block has size $s$, the first index with $d_s = d_{s+1}$, which is also the exponent of $t - \lambda$ in $\mu_T$.

Definition Let $A \in \mathbb{F}^{n \times n}$ have a characteristic polynomial that splits over $\mathbb{F}$. Work one eigenvalue at a time. For each eigenvalue $\lambda$, with $N = A - \lambda I$,

1. compute $d_k = n - \operatorname{rank} N^k$ until it stops growing, and read off the block sizes,
2. find the spaces $E_k = \ker N \cap \operatorname{im} N^{k-1}$, choose a basis of the top nonzero level and extend it level by level down to $E_1$, keeping the vectors already chosen, and for each vector $b$ first added at level $k$ solve $N^{k-1}w = b$,
3. list each chain $(N^{k-1}w, \dots, Nw, w)$ bottom first as consecutive columns of $P$.

Every step may use $A - \lambda I$ on all of $\mathbb{F}^n$, because it is invertible on the generalized eigenspaces of the other eigenvalues, so the kernels, the spaces $E_k$ and the tops $w$ automatically lie in $G_\lambda$. With the chains of all eigenvalues placed side by side, $P$ is invertible and $P^{-1}AP = J$. To check a proposed answer without computing $P^{-1}$, verify that $\det P \ne 0$ and that $AP = PJ$ column by column. That means $Ap = \lambda p$ for the first column $p$ of each chain and $Ap_j = \lambda p_j + p_{j-1}$ for each later column $p_j$ of a chain.

Definition Let $T \colon V \to V$ be linear on a finite-dimensional space over $\mathbb{F}$ with split characteristic polynomial. Then all Jordan forms of $T$ contain the same blocks, counted with multiplicity, and differ only in the order of the blocks. Precisely, for every eigenvalue $\lambda$ and every $k \ge 1$, each Jordan form of $T$ has exactly

blocks equal to $J_k(\lambda)$, where $d_k = \dim \ker(T - \lambda I)^k$. To prove it, let $A$ be a matrix of $T$ and $J = P^{-1}AP$ any Jordan form. For every $k$,

and $x \mapsto Px$ maps $\ker(J - \lambda I)^k$ one-to-one onto $\ker(A - \lambda I)^k$, so $J$ and $A$ have the same numbers $d_k$. Applied to the Jordan matrix $J$, the counting rule turns these numbers into the block counts of $J$, and the numbers depend only on $T$. Consequently two matrices in $\mathbb{F}^{n \times n}$ with split characteristic polynomials are similar over $\mathbb{F}$ exactly when their Jordan forms have the same blocks, since reordering the blocks is itself a similarity by a permutation matrix. When $V \ne 0$ the Jordan basis is not unique. Scaling a whole chain by a scalar $c \notin \{0, 1\}$, lifting the bottom of a chain of length at least $2$ to a different top, or reordering two or more chains gives another Jordan basis.

Definition The Jordan form is not a continuous function of the entries. An arbitrarily small change in a matrix can change its block sizes, and when that happens the eigenvalues can move much farther than the entries did. A standard example is the real family

For $\varepsilon = 0$ the Jordan form is the single block $J_2(1)$. For $\varepsilon > 0$ the eigenvalues $1 \pm \sqrt{\varepsilon}$ are distinct, with eigenvectors $(1, \sqrt{\varepsilon})$ and $(1, -\sqrt{\varepsilon})$, so $A_\varepsilon$ is diagonalizable over $\mathbb{R}$, with two blocks of size $1$. For $\varepsilon < 0$ the characteristic polynomial has no real roots, so $A_\varepsilon$ is not similar over $\mathbb{R}$ to any Jordan matrix, while over $\mathbb{C}$ the eigenvalues $1 \pm i\sqrt{-\varepsilon}$ are distinct and $A_\varepsilon$ is diagonalizable. The entries move by $\lvert \varepsilon \rvert$ but the eigenvalues move by $\sqrt{\lvert \varepsilon \rvert}$. For $\varepsilon > 0$ the angle between the two eigenvectors above is $2\arctan\sqrt{\varepsilon}$, which tends to $0$ as $\varepsilon \to 0$, so the diagonalizing basis becomes nearly dependent.

Exact Jordan structure rests on exact zero tests, such as whether $A_\varepsilon - I$ has a nonzero kernel, which happens only for $\varepsilon = 0$. The singular values of $A_\varepsilon - I$ are $1$ and $\lvert \varepsilon \rvert$. A tiny computed singular value is consistent with a kernel of dimension $0$ and with a kernel of dimension $1$, so a rank decision from rounded or measured data needs a stated tolerance and an error model. For numerical work a unitary Schur form $A = QRQ^{*}$, with $Q$ unitary and $R$ upper triangular, is usually a better starting point than a Jordan basis or a nearly dependent eigenvector basis, because a unitary change of basis preserves lengths and so does not magnify errors.