# Does the Jordan form of the matrix depend on where the Jordan blocks are placed?

Let's say that we have the matrix $$\begin{equation*} A = \begin{pmatrix} 1 & -1 & 0 \\ -1 & 4 & -1 \\ -4 & 13 & -3 \end{pmatrix}. \end{equation*}$$ Now, the characteristic polynomial denoted by $$C_p(A)= \lambda(\lambda-1)^2$$ implies that there will be one jordan block corresponding to the eigenvalue $$\lambda=0$$. Hence all that is left to find are the number of jordan blocks for $$\lambda=1$$. After row reducing it is clear that the eigenspace for $$\lambda=1$$ is one dimensional and so it must be a jordan block of size $$2$$.

So here is my question. Does the order of the Jordan blocks matter? That is, are $$J_2(1)\oplus J_1(0)$$ and $$J_1(0)\oplus J_2(1)$$ equivalent expressions? Or is there some order that must be adhered to?

Thanks for any help.

• No there is no specific order required, just as if you diagonalize a matrix, there is no real reason to pick one ordering of eigenvalues over another. – peek-a-boo Aug 19 at 23:12
• Jordan normal form is only defined up to reordering. Also, you can conjugate your block-diagonal matrix by a permutation matrix to permute the blocks arbitrarily. – David A. Craven Aug 19 at 23:14
• The Jordan canonical form (when it exists) is unique up to permutation of the blocks, so any order you pick is up to you. As a comparison, the fundamental theorem of arithmetic roughly says every there is a unique prime factorization (unique up to ordering of the factors). So, for example, $40 = 2^3 \times 5 = 2\times 5 \times 2^2 = 5 \times 2^3$. Would you really care which order it is presented in? – peek-a-boo Aug 19 at 23:17
• @peek-a-boo That makes a lot of sense. Thanks for clearing this up. – John Kameas Aug 19 at 23:21
• In some countries the Jordan form is taught with the extra 1's below the main diagonal. This is fine; the fact that the two ways are visibly "similar" is one proof that any matrix is similar to its transpose. – Will Jagy Aug 19 at 23:50