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

Question

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.

Answer 1

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