A Jordan Block is defined to be a square matrix of the form:

$\left[\begin{array}{ccccc}\mathrm{\λ}& 1& 0& \cdots & 0\\ 0& \mathrm{\λ}& 1& \cdots & 0\\ ⋮& ⋮& \ddots & & ⋮\\ 0& 0& 0& \mathrm{\λ}& 1\\ 0& 0& 0& \cdots & \mathrm{\λ}\end{array}\right]$

for some scalar l.

A Jordan Matrix is a matrix that has Jordan Blocks on its diagonal and the rest of the entries equal to 0:

$\left[\begin{array}{cccccccccc}\mathrm{\λ\_\_1}& 1& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& \ddots & 1& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& \mathrm{\λ\_\_1}& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& \mathrm{\λ\_\_2}& 1& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& \ddots & 1& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& \mathrm{\λ\_\_2}& & 0& 0& 0\\ 0& 0& 0& 0& 0& 0& \ddots & 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& \mathrm{\λ\_\_n}& 1& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& \ddots & 1\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& \mathrm{\λ\_\_n}\end{array}\right]$

where the colored regions are the Jordan Blocks of the matrix. Furthermore, note that the $\mathrm{\λ\_\_n}$ values in each Jordan block need not to be all equal.

Any square matrix M is similar to a Jordan matrix  J, which is called the Jordan Canonical Form of M. For M, There exists an invertible Q such that:

M = Q · J · $Q{}^{\mathit{-}\mathit{1}}$

As we can observe, J is upper triangular and almost diagonal, from this we know that M and J have the same:

As such, the calculations for these common features are simplified by working with the Jordan Canonical Form of M instead of M.

First, enter the matrix you wish to reduce to Jordan Canonical Form below and click the "Set Matrix" button:

	Step 1: Eigenvalues and Multiplicities
	We will calculate the eigenvalues of the matrix by finding the matrix's characteristic polynomial. Thus, we solve:   The characteristic polynomial for the matrix is:   This gives eigenvalues  with multiplicities of , where the left side of each equation is the eigenvalue and the right side of each equation is the multiplicity of that eigenvalue.

Step 2: Eigenvalues

Here is the explanation of the step that is applied to each specific eigenvalue. Mathematical Process for Finding the Jordan Canonical Form Let the eigenvalues of the matrix M be ${\mathrm{\λ}}_1\, ..\.\, \mathrm{\λ\_\_n}$ with multiplicities $\mathrm{m\_\_1}\, \mathrm{m\_\_2}\, ..\.\, \mathrm{m\_\_n}$. The process is the following: 1.  Fix l = ${\mathrm{\lambda }}_i$ and compute $d_1$ = Nullity(M - ${\mathrm{\lambda }}_{}$). Note that this is step is basically computing the dimension of the eigenspace of M.  Furthermore, $d_1$ is the number of Jordan blocks that our Jordan matrix will have. 2.  Compute $d_2$ = Nullity${\left(M - {\mathrm{\lambda }}_{}\cdot I\right)}^2$. Here, $d_2$- $d_1$ is the number of Jordan blocks of size at least 2x2 that our Jordan matrix will have. 3.  Compute $d_3$ = Nullity${\left(M - \mathrm{\λ}\cdot I\right)}^3$. Here, $d_3$- $d_2$ is the number of Jordan blocks of size at least 3x3 that our Jordan matrix will have. 4.  Continue doing the same process until $\mathrm{d\_\_k} \= \mathrm{m\_\_i}$. 5.  Put together logically the quantities of Jordan blocks that were obtained in the previous steps to obtain the number and size of Jordan blocks of the given eigenvalue.   Note: There are many ways to calculate the nullity requested above:   •  One of the simplest ways to do so is by transforming the matrix into row echelon form and counting the number of rows at the end that have all entries equal to zero. •  Another way to do it is by solving the equation ${\left(M - {\mathrm{\lambda }}_{}\cdot I\right)}^k\cdot v \= 0$ for v, finding a basis for the set of solutions and then counting the number of elements this basis has.   For a computational example using the entered matrix, see the below section.   Some Maple Help for this Step Choose the eigenvalue l = Choose23 that you wish to calculate for in this step. Below, you can walk through the process by calculating it for k and increasing k by one in each iteration. You can start at 1 by pressing the "Reset Step" button and calculate for the next degree by pressing the "Calculate for k + 1" button.   The current degree of ${\left(M - \mathrm{\λ}\cdot I\right)}^k$ is $k$ = The eigenvalue has multiplicity $m$ = . 1.  Write ${\left(M - \mathrm{\lambda }\cdot I\right)}^k$ in Reduced Row Echelon Form. → 2.  Count the number of rows with all entries equal to zero which will give Nullity${\left(M - \mathrm{\λ}\cdot I\right)}^k$. In this case, the dimension is . 3.  Calculate $\dim {\left(M - \mathrm{\λ}\cdot I\right)}^k - \dim$${\left(M - \mathrm{\λ}\cdot I\right)}^{k-1}$. In this case, we get  so there is(are)  Jordan block(s) of size at least  by . 4.  Since is    then we  repeat this step for k + 1 .

	Step 3: Repeat
	This step is computing Step 2 for each of the eigenvalues. After computing Step 2 for each of the eigenvalues and logically considering all obtained number of Jordan blocks. We now write the Jordan Blocks diagonally in our new matrix by putting the Jordan blocks of each eigenvalue adjacent to one another.

Putting everything together the Jordan matrix of M is .
Furthermore, the transition matrix Q in this case is .

Diagonalizable Matrices

Note that, if the characteristic polynomial of the matrix splits and each eigenvalue has multiplicity 1 then the Jordan Canonical Form of the matrix will be a diagonal matrix.

This is because on the first iteration of step 2 applied to eigenvalue l, we would have that the nullity of the eigenspace is 1 and the multiplicity for the root is 1.

Therefore, no more steps are needed and so the only Jordan block for the eigenvalue will be the 1 x 1 matrix: [l].