There were many questions about Jordan normal forms, so here is an e-mail trying to explain it a little better. ---A Jordan block is a matrix with a constant number c down the diagonal and 1s right above the diagonal. examples: 2 1 0 , (3) , 1 1 0 2 1 0 1 0 0 2 In particular, a diagonal entry is a Jordan block. The matrix 1 1 0 2 is NOT a Jordan block because the diagonal entries are not the same. The matrix 1 0 1 0 1 0 0 0 1 is NOT a Jordan block becuase the one is not directly above the diagonal. ---Jordan normal form: A matrix is in Jordan normal form if it consists of Jordan blocks down the diagonal. Examples 1 0 , 1 1 0 0 , in particular all diagonal matrices are 0 2 0 1 0 0 in Jordan normal form. 0 0 2 1 0 0 0 2 NOT in Jordan normal form: 1 1 , 0 0 1 0 2 1 1 0 0 1 0 The Jordan blocks in the matrix 1 1 0 0 0 consist of the three blocks (1) , (2) and 1 1 0 0 1 1 0 0 0 1 1 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 0 0 2 The MAIN THEOREM: Every nxn matrix is similar to a UNIQUE matrix in Jordan normal form (over the complex numbers and upto reordering of the Jordan blocks) Applications: 1. Are the matrices 1 1 0 0 and 1 1 0 0 0 1 1 0 0 1 0 0 0 0 1 0 0 0 1 1 0 0 0 1 0 0 0 1 similar? If we did not know the theorem, it would be hard to answer this question because both these matrices have the eigenvalue 1 with algebraic multiplicity 4 and geometric multiplicity 2. However, using the theorem we can see that these matrices are not similar. They are both in Jordan normal form. The theorem asserts that the Jordan normal form is unique. But the blocks in these two matrices are different The first matrix has the blocks (1) and 1 1 0 where as the second 0 1 1 0 0 1 matrix has two blocks of type 1 1 0 1 , so they are not similar. Application 2: If A has characteristic polynomial (x-3)^3 (x-4), what are the possible Jordan normal forms of A? The block containing the eigenvalue 4 has to be the diagonal block (4) The blocks containing the eigenvalue 3 have a few possibilities. 1. There can be 3 blocks, i.e. the matrix is diagonalizable 3 0 0 0 0 3 0 0 0 0 3 0 0 0 0 4 2. There can be 2 blocks containing the eigenvalue 3 3 1 0 0 0 3 0 0 0 0 3 0 0 0 0 4 3. There can be one block containing the eigenvalue 3 3 1 0 0 0 3 1 0 0 0 3 0 0 0 0 4 These are all the possibilities. How about 3 0 0 0 0 3 1 0 0 0 3 0 0 0 0 4 ? This is the same as case 2. The Jordan blocks have been permuted. This matrix is similar to the matrix occuring in case 2. This is what I mean by unique up to reordering of the blocks. I hope this helps. E-mail me if you still have questions and let me know if there are any typos. Izzet