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Jordan Normal Form Matrix

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

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Jordan Normal Form Matrix -

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# Jordan Normal Form Matrix

## Jordan Matrix Calculator

### How to calculate the Jordan Normal Form for a matrix?

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 $M = S . J . \bar{S}$

### What is Jordan Decomposition?

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

### How to calculate a power of a Jordan matrix?

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

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