Triangularization and Jordan Normal Form

When diagonalization is not possible

Diagonalization requires a basis consisting entirely of eigenvectors. Such a basis does not always exist. A weaker result is still available: if the characteristic polynomial of $A \in \mathbb{K}^{n \times n}$ splits into linear factors over $\mathbb{K}$ , then an invertible matrix $S$ transforms $A$ into an upper triangular matrix.

R=S^{-1}AS

The diagonal entries are the eigenvalues, repeated according to their algebraic multiplicities. Over $\mathbb{C}$ , every characteristic polynomial splits. Over $\mathbb{R}$ , this condition can fail when nonreal eigenvalues occur.

Jordan normal form sharpens this triangular description. It records the eigenvalues and shows exactly $how many eigenvectors are missing from a basis$ .

One Jordan block

A Jordan block of size $s$ for the eigenvalue $\lambda$ is

J_s(\lambda)=\begin{pmatrix}\lambda&1&0&\cdots&0\\0&\lambda&1&\ddots&\vdots\\\vdots&\ddots&\ddots&\ddots&0\\0&\cdots&0&\lambda&1\\0&\cdots&\cdots&0&\lambda\end{pmatrix}

The $nilpotent matrix$ $N_s$ has ones on the superdiagonal and zeros elsewhere. A block of size $1$ is simply $(\lambda)$ . A larger block signals that one eigenvector must be extended by generalized eigenvectors.

Generalized eigenvector chains

Vectors $v_1,\ldots,v_s$ form a Jordan chain for $\lambda$ when

(A-\lambda I)v_1=0

The $first vector$ is an ordinary eigenvector. We require $v_1\neq0$ . The recurrence then forces the chain vectors to be linearly independent. Each following vector maps to the previous one after applying $A-\lambda I$ . In the ordered basis $(v_1,\ldots,v_s)$ , the restriction of $A$ to this chain is represented by $J_s(\lambda)$ .

The complete Jordan form

Suppose the characteristic polynomial of $A$ has the following form.

\chi_A(t)=\prod_{i=1}^{k}(\lambda_i-t)^{a_i}

If this polynomial splits over $\mathbb{K}$ with distinct eigenvalues $\lambda_1,\ldots,\lambda_k$ , there is an invertible matrix $S$ giving the following block-diagonal form. Group blocks associated with the same eigenvalue:

\begin{aligned} &D_i=\operatorname{diag}\bigl(J_{s_{i1}}(\lambda_i), \\ &\ldots,J_{s_{iq_i}}(\lambda_i)\bigr) \\ &i=1,\ldots,k \end{aligned}

For each eigenvalue, the block sizes satisfy

s_{i1}+\cdots+s_{iq_i}=a_i

The $number of blocks$ for $\lambda_i$ equals the geometric multiplicity $\dim\ker(A-\lambda_iI)$ . The block sizes are uniquely determined by $A$ , although their order along the diagonal is not.

Question	What the Jordan blocks show
Is $A$ diagonalizable?	Yes exactly when every Jordan block has size $1$ .
How many independent eigenvectors belong to $\lambda_i$ ?	One for each Jordan block associated with $\lambda_i$ .
Why is a repeated eigenvalue not enough to decide diagonalizability?	The same algebraic multiplicity can be split among blocks of different sizes.

Reading Jordan Blocks from a Matrix

Consider

A=\begin{pmatrix}1&1&0\\0&1&0\\0&0&2\end{pmatrix}=J_2(1)\oplus J_1(2)

Its characteristic polynomial is

\chi_A(t)=(1-t)^2(2-t)

For $\lambda=1$ ,

A-I=\begin{pmatrix}0&1&0\\0&0&0\\0&0&1\end{pmatrix}

The eigenvalue $1$ has algebraic multiplicity $2$ but $only one independent eigenvector$ . Therefore $A$ is not diagonalizable. Consider the following two vectors.

v_1=\begin{pmatrix}1\\0\\0\end{pmatrix},\qquad v_2=\begin{pmatrix}0\\1\\0\end{pmatrix}

The two vectors form a Jordan chain because $(A-I)v_1=0$ and $(A-I)v_2=v_1$ . The eigenvalue $2$ contributes the separate block $J_1(2)$ .

Matrix Powers from Jordan Blocks

Writing $J_s(\lambda)=\lambda I+N$ with $N^s=0$ makes powers explicit:

For integers $m\ge0$ and $\lambda\neq0$ :

J_s(\lambda)^m=\sum_{j=0}^{\min(m,s-1)}\binom{m}{j}\lambda^{m-j}N^j

When $\lambda=0$ , use $J_s(0)^m=N^m$ directly, with $N^0=I$ . For example, the nonzero superdiagonal accumulates with each multiplication:

\begin{pmatrix}1&1\\0&1\end{pmatrix}^2=\begin{pmatrix}1&2\\0&1\end{pmatrix},\qquad \begin{pmatrix}1&1\\0&1\end{pmatrix}^3=\begin{pmatrix}1&3\\0&1\end{pmatrix}

This formula explains why powers of a non-diagonalizable matrix contain $polynomial factors$ in $m$ in addition to powers of the eigenvalues. It gives the exact matrix powers used to analyze recurrences and differential equations.

Jordan form is primarily an exact algebraic classification. It is highly sensitive to small numerical perturbations, so numerical software usually computes a Schur decomposition for stable calculations.