Triangularization and Jordan Normal Form

Matrix Triangularization

Even though matrix $A$ cannot be diagonalized, the following theorem still holds.

For matrix $A \in \mathbb{K}^{n \times n}$, if the characteristic polynomial of $A$ can be factored into linear factors, then $A$ is similar to an upper triangular matrix $R \in \mathbb{K}^{n \times n}$. The entries on the diagonal of this upper triangular matrix are all the eigenvalues of $A$.

This similarity relationship is expressed as:

where $S \in \mathbb{K}^{n \times n}$ is an invertible matrix.

This upper triangular form can be described more precisely through the Jordan Normal Form.

Jordan Normal Form

Let $A \in \mathbb{K}^{n \times n}$ with characteristic polynomial:

with distinct eigenvalues $\lambda_1, \ldots, \lambda_k \in \mathbb{K}$. Then there exists an invertible matrix $S \in \mathbb{K}^{n \times n}$ such that:

This form shows how a matrix can be organized into simpler blocks, where each block is associated with one particular eigenvalue.

Jordan Block Structure

For each $i = 1, \ldots, k$, the Jordan block $\lambda_i \cdot I_{r_i} + N_i$ has a very distinctive structure. Imagine it like an almost perfect staircase, where each step has the same value (namely the eigenvalue $\lambda_i$), but there are "connectors" in the form of the number 1 at certain positions that make this structure unique.

In this structure, the eigenvalue $\lambda_i$ dominates the main diagonal, while the number 1 appears at certain positions above the diagonal (called the superdiagonal). The positions of 0 and 1 on the superdiagonal determine how the Jordan block is divided into smaller sub-blocks. This structure provides complete information about how the linear transformation operates on the vector space associated with that eigenvalue.