Jordan Decomposition

The process of decomposing a square matrix $A$ into this form:

Here:

• $J$ is the Jordan form of $A$
• $P$ is an invertible matrix where the columns are generalized eigenvectors of $A$

Generalizes diagonalization when matrix is not diagonalizable. Every square matrix $A$ is Jordan-decomposable.

Jordan Block

Denoted by $B(\lambda,m)$. An upper triangular matrix $B_{m\times m}$ is a Jordan block iff:

• All $m$ diagonal elements are the same eigenvalue ($\lambda$) and
• All super-diagonal elements are $1$

Jordan Form

Aka. Jordan canonical form. $J$ from the decomposition.

Here:

• $\lambda_i$ is an eigenvalue of $A$ for $i=1,2,\dots,m$
• $m_i \in \mathbb{R}$ where $\sum m_i = n$

Unique for a given matrix, up to the reordering of the Jordan blocks.

2 matrices $A$ and $B$ are similar iff they have the same Jordan form (up to the reordering of the Jordan blocks).

TODO: explain how jordan decomposition is done easily.

Relationship between multiplicities

Suppose Jordan blocks of the eigenvalue $\lambda$ are of the sizes $k_1,k_2,\dots,k_r \ge 1$:

• $g_\lambda = r$ Number of Jordan blocks associated with $\lambda$ is equal to the geometric multiplicity of $\lambda$.
• $a_λ = \sum_{i=1}^{r} k_i \ge r$ Algebraic multiplicity of $\lambda$ is equal to the sum of sizes of the Jordan blocks. And it’s greater than the number of Jordan blocks. Obvious.
• $m_\lambda = \text{max}(k_1, k_2,\dots, k_r)$
• $g_\lambda (m_\lambda - 1) \ge a_\lambda - g_\lambda = \sum_{i=1}^{r} {k_i - 1} \ge m_\lambda - 1$
• $a_\lambda-g_\lambda = m_\lambda-1$ if the size of the matrix $n\le3$