Jordan Decomposition | Sahithyan's Notes

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

A = PJP^{-1}

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:

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

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

J = \text{diag}(B(λ_1, m_1), B(λ_2, m_2), \dots , B(λ_k, m_k))

Here:

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).

Step 1: Find the characteristic polynomial $f(\lambda) = \det(\lambda I - A)$ .

Step 2: Solve $f(\lambda) = 0$ . Each distinct root is an eigenvalue. The power of $(\lambda - \lambda_i)$ in $f$ is $a_{\lambda_i}$ .

Step 3: For each eigenvalue $\lambda_i$ , compute:

$g_{\lambda_i} = n - \text{rank}(A - \lambda_i I)$
Number of Jordan blocks for $\lambda_i$ .
$a_{\lambda_i}$
Sum of sizes of all blocks for $\lambda_i$ .

Step 4: Determine the sizes of the $g_\lambda$ blocks, which must sum to $a_\lambda$ .

If $a_\lambda = g_\lambda$ : all blocks are $1 \times 1$ .

Otherwise, compute $r_k = \text{rank}(A - \lambda I)^k$ for $k = 0, 1, 2, \dots$

Number of blocks of size $\ge k$ is:

n_{\ge k} = r_{k-1} - r_k

Number of blocks of size exactly $k$ is:

n_k = r_{k-1} - 2r_k + r_{k+1}

Step 5: Assemble $J$ by placing each Jordan block $B(\lambda_i, k)$ along the diagonal.

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 $n\le3$