Jordan Decomposition
The goal of a Jordan decomposition is to diagonalize a given square matrix. If there is an invertible n×n matrix C and a diagonal matrix D such that A=CDC-1, then an n×n matrix A is diagonalizable.
Procedure-
Choose a square matrix (m X m) (e.g., 3 X 3, 4 X 4, 5 X 5, etc.,)
Otherwise-Pop up error – select number of rows and Columns should be same (or matrix dimension mismatched)
For a given matrix,
A = $\begin{bmatrix} 2 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 2 \end{bmatrix}$
The aim of Jordan decomposition is to diagonalize a given square matrix A, if A=PDP-1 is possible, where P is an invertible matrix and D is diagonal matrix. We'll go into the specifics of how matrix P and matrix D are formed later. Matrix P and D are derived from matrix A.
Firstly, we’ll find the eigen values of the matrix A
| A – λ*I | = 0 (I = identity matrix)
Or, $\begin{bmatrix} 2 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 2 \end{bmatrix} -$ $\lambda*\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ = 0
Or, $\begin{bmatrix} \mathbf{(2 -}\lambda) & \mathbf{1} & \mathbf{0} \\ \mathbf{1} & \mathbf{(2 -}\lambda) & \mathbf{1} \\ \mathbf{0} & \mathbf{1} & \mathbf{(2 -}\lambda) \end{bmatrix}$ = 0
Or, (2-λ) ((2-λ) × (2-λ)-1×1)-1(1× (2-λ)-1×0) +0(1×1-(2-λ) ×0) = 0
Or,(2-λ)((4-4λ+λ2)-1)-1((2-λ)-0)+0(1-0)=0
Or,(2-λ)(3-4λ+λ2)-1(2-λ)+0(1)=0
Or,(6-11λ+6λ2-λ3)-(2-λ)+0=0
Or,(-λ3+6λ2-10λ+4)=0
Or,-(λ-2)(λ-0.5858)(λ-3.4142)=0
Or,(λ-2)=0or(λ-0.5858)=0or(λ-3.4142)=0
So, The eigenvalues of the matrix A are given
by λ=0.5858,2,3.4142
You can apply Newton Raphson method to find a good approximation for the root of a real-valued function. You can use this method here to find the eigen values (or, λ’s)
Please read through the matrix's minor and co-factor in to understand the finding of the determinant value in step 3. I've already written an article regarding minors of a matrix.
Now, calculate the eigen vectors from the corresponding eigen values.
In our case, eigen values are 0.5858, 2, 3.4142
For, λ = 0.5858
A - λ * I
= A - 0.5858 * I
= $\begin{bmatrix} 2 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 2 \end{bmatrix}$ – 0.5858*$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
=
 = |
1.4142 | 1 |  0 |
|
|---|---|---|---|---|
| = | 1 | 1.4142 | 1 | |
| 0 | 1 | 1.4142 |
Now, do row operations to reduce the matrix
Now, reduce this matrix
R1←R1÷1.4142
| = |
|
|---|
R2←R2-R1
| = |
|
|---|
Interchanging rows R2↔︎R3
| = |
|
|---|
R1←R1-0.7071×R2
| = |
|
|---|
R3←R3 - 0.7071×R2
| = |
|
|---|
One can calculate row echelon form to reduce a matrix
Now, compute
A*x - λI*x = 0
Or, (A - λI)x = 0
Or, (A - 0.5858 * I)x = 0
Or, $\begin{bmatrix} 1 & 0 & - 1 \\ 0 & 1 & 1.4142 \\ 0 & 0 & 0 \end{bmatrix}*$ $\begin{bmatrix} x1 \\ x2 \\ x3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$
Or, x1-x3=0, x2+1.4142x3=0
Or, x1=x3, x2=-1.4142x3
Now, for eigen value, λ = 0.5858, corresponding eigen vector is
v1= $\begin{bmatrix} x3 \\ - 1.4142x3 \\ x3 \end{bmatrix}$
let x3 = 1
v1= $\begin{bmatrix} 1 \\ - 1.4142 \\ 1 \end{bmatrix}$
We found the eigen vector for the eigen value, =0.5858, only in step 4 above. The same method may be used to calculate the eigen vectors for λ=2 and 3.4142.
Corresponding eigen vectors for eigen values 2 & 3.4142 are
v2= $\begin{bmatrix} - 1 \\ 0 \\ 1 \end{bmatrix}$
and
v3= $\begin{bmatrix} 1 \\ 1.4142 \\ 1 \end{bmatrix}$ respectively.
To allow diagonalization, the number of eigenvectors must be equal the given square matrix’s dimension.
If the number of eigenvalues is less than the dimension of the given square matrix, a matrix cannot be diagonalized and show pop-up error.
Pop up error – ‘not diagonalizable!’
Now, initialize the P matrix. P matrix columns are formed from the eigen vectors derived from the eigen values 0.5858, 2, and 3.4142 or they are
[v1 v2 v3]
or, P = $\left\lbrack - \begin{matrix} 1 & - 1 & 1 \\ 1.4142 & 0 & 1.4142 \\ 1 & 1 & 1 \end{matrix} \right\rbrack$
The diagonal matrix (D) of the above matrix A contains the eigen values of matrix A as the following diagonal elements:
D = $\begin{bmatrix} 0.5858 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3.4142 \end{bmatrix}$
8. Now the final step is to check whether the matrix P is invertible or not. If matrix P is not invertible then display the pop-up notification
Pop up error – ‘not diagonalizable!’
The values of matrices A, P, and P-1 will only be displayed if matrix P is invertible.