Tool to calculate the Jordan Normal Form of a Matrix (by Jordan reduction of a square matrix) to get, by decomposition, 2 matrices S and J such that M = S . J . S̄

Jordan Normal Form Matrix - dCode

Tag(s) : Matrix

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Take $ M $ a square matrix of size $ n $, which has for eigen values the set of $ \lambda_n $.

__Example:__ $$ M = \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 1 & 2 \end{bmatrix} \Rightarrow \lambda_n = \begin{pmatrix} 3 \\ 3 \\ 3 \end{pmatrix} $$

A matrix $ M $ of size $ n \times n $ is diagonalizable if and only if the sum of the dimensions of its eigen spaces is $ n $.

If $ M $ is not diagonalisable, there exists an almost diagonal matrix $ J $, called Jordan Normal Form, of the form $$ \begin{bmatrix} \lambda_i & 1 & \; & \; \\ \; & \lambda_i & \ddots & \; \\ \; & \; & \ddots & 1 \\ \; & \; & \; & \lambda_i \end{bmatrix} $$

__Example:__ Here, $ M $ has only 2 eigen vectors : $ v_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} $ et $ v_2 = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} $, so is not diagonalizable, but has for **Jordan matrix** (canonical form) $$ M=\begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 1 \\ 0 & 0 & 3 \end{bmatrix} $$

__Example:__ Alternative method: calculate the matrix $ S $ by finding a third vector $ v_3 $ such as $ (M - 3 I_3) v_3 = k_1 v_1 + k_2 v_2 \Rightarrow v_3 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} $. So $$ S = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix} $$ and $ M = S . J . \bar{S} $

Jordan's decomposition is obtaining, from a matrix $ M $, the matrices $ S $ and $ J $ such that $ M = S . J . \bar {S} $

If $ M = SJS^{-1} $ Then $ M^k = SJ^kS^{-1} $ (see matrix powers).

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