Tool to calculate the Jordan Normal Form of a Matrix (by Jordan reduction of a square matrix). The Jordan matrix is used in analysis, from a matrix M, the Jordan decomposition provides 2 matrices S and J such that \( M = S. J. \bar{S} \).

Jordan Normal Form Matrix - dCode

Tag(s) : Matrix

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Tool to calculate the Jordan Normal Form of a Matrix (by Jordan reduction of a square matrix). The Jordan matrix is used in analysis, from a matrix M, the Jordan decomposition provides 2 matrices S and J such that \( M = S. J. \bar{S} \).

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 \( M = S . J . \bar{S} \)

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

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