Find the Jordan reduction of a 3×3 matrix:

This is the second of the two matrices returned by JordanDecomposition:

Determine the algebraic and geometric multiplicities of the eigenvalues of the following matrix:

From the diagonal elements of the Jordan form, there is a single eigenvalue of algebraic multiplicity 4:

As there are 1s all along the superdiagonal, there is a single subblock and thus the geometric multiplicity is 1:

Eigensystem confirms that there is a single linearly independent eigenvector:

The algebraic multiplicity of an eigenvalue equals the length of the diagonal of block of the eigenvalue in the Jordan form, and the geometric multiplicity equals the number of subblocks delineated by a 0 in the superdiagonal in each block. Find the algebraic and geometric multiplicities of the eigenvalues of the following matrix:

From the diagonal elements of the Jordan form, 3 and 4 have algebraic multiplicity 2, and 5 has algebraic multiplicity 3:

From the full matrix it can be seen that the 4 has geometric multiplicity 1, and 3 and 5 have multiplicity 2:

The subblocks can also be counted using BlockDiagonalMatrix; 3 has two bocks, 4 a single , and 5 a and a block:

A square matrix has a complete set of eigenvectors, and thus is diagonalizable, iff its matrix is diagonal:

Test if a particular matrix is diagonalizable:

Confirm using DiagonalizableMatrixQ:

Estimate the probability that a 4×4 matrix of ones and zeros will be diagonalizable:

A matrix m₁ and a random invertible matrix s give a similar matrix m₂:

The Jordan reduction of m₁:

The similar matrix m₂ has the same reduction:

JordanReduce[m] is equivalent to the second matrix returned by JordanDecomposition[m]:

Compute the Jordan form of an integer matrix:

It is notably slower to compute the full decomposition:

Check that the diagonals agree:

A matrix is diagonalizable iff its Jordan reduction is diagonal:

EigenvalueDecomposition[m] exists if and only if m's Jordan reduction is diagonal:

The Jordan form of nd has a 1 above the diagonal:

Here the decomposition exists:

As expected, the Jordan matrix is diagonal in this case:

For a real symmetric numerical matrix, the Jordan form is diagonal and real valued:

For a real antisymmetric numerical matrix, the Jordan form is diagonal with pure imaginary diagonal entries:

For a unitary numerical matrix, the Jordan form is diagonal:

The diagonal entries lie on the unit circle:

For a normal numerical matrix, the Jordan form is diagonal:

SchurDecomposition[n,RealBlockDiagonalFormFalse] for a numerical normal matrix :

This is the same matrix one constructs from the eigenvalues:

This coincides with the Jordan form up to reordering:

The matrix m has some machine-precision entries:

Due to numerical rounding, the deficient eigenspace at 2. is split into two separate eigenspaces:

Perform the computation using exact arithmetic to see that the matrix is not diagonalizable (having a 1 on the superdiagonal from the eigenvalue 2 with multiplicity of two):

Exponentiate the numeric Jordan form, with its diagonal entries reordered to match the exact decomposition:

Comparing numerically to exponentiating the exact Jordan form shows a considerable difference in the second column of the first row: