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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.

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    $\begingroup$ 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. $\endgroup$
    – peek-a-boo
    Commented Aug 19, 2020 at 23:12
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    $\begingroup$ 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. $\endgroup$ Commented Aug 19, 2020 at 23:14
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    $\begingroup$ 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? $\endgroup$
    – peek-a-boo
    Commented Aug 19, 2020 at 23:17
  • $\begingroup$ @peek-a-boo That makes a lot of sense. Thanks for clearing this up. $\endgroup$
    – Alias K
    Commented Aug 19, 2020 at 23:21
  • $\begingroup$ 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. $\endgroup$
    – Will Jagy
    Commented Aug 19, 2020 at 23:50

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The order does not matter. Any twoJordan forms of a matrix are similar and different order of blocks gives a smilar Jordan form.

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