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Suppose I have a matrix $A \in M_{n \times n}(\mathbb{C})$ such that its minimal polynomial is either $x-1$ or $(x-1)^{2}$. What are its possible Jordan Canonical Forms? I was thinking that if its minimal polynomial is $x-1$, then its Jordan canonical form is $I_{n}$, the $n \times n$ identity matrix. But if its minimal polynomial is $(x-1)^{2}$ then the number of its Jordan Canonical Forms depend on $n$. I was thinking that the number of forms is $\lfloor \frac{n}{2} \rfloor$. For example, when $n = 7$, we have that $V \cong \left( \mathbb{C}[x] / (x-1) \right)^{5} \oplus \mathbb{C}[x] / (x-1)^{2}$, or $V \cong \left( \mathbb{C}[x] / (x-1) \right)^{2} \oplus \left( \mathbb{C}[x] / (x-1)^{2} \right)^{2}$ or $V \cong \mathbb{C}[x] / (x-1) \oplus \left( \mathbb{C}[x] / (x-1)^{2} \right)^{3}$, which gives $3$ distinct Jordan forms. Also, are the matrices $\begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\0 & 0 & 0 & 1 \end{bmatrix}$ and $\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\0 & 0 & 0 & 1 \end{bmatrix}$ considered different Jordan canonical forms or the same for $n=4$ and the minimal polynomial $(x-1)^{2}$.

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    $\begingroup$ They are considered the same canonical form. The Jordan form is unique up to the order of the blocks. The rest of your computations seem to be correct. $\endgroup$ – Arturo Magidin Dec 11 '19 at 17:03
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Consider the special case when $A$ is nilpotent:

If minimal polynomial has degree $m=1$, then largest Jordan block will be of size $1\times 1$, thus JCF is diagonal. This can be generalized to any matrix.

If minimal polynomial has degree $m=2$, then largest Jordan block will be of size $2\times 2$, and the rest of the Jordan blocks will depend on the dimensions of $N(A^i)$ (null space of $A^i$), larger the $n$ the more options we have. For example, when $n=7$, we can have JCF with jordan blocks $(2,2,2,1),$ $(2,2,1,1,1),$ $(2,1,1,1,1,1),$ assuming that the Jordan blocks are used from largest to smallest, otherwise we can change the position of jordan blocks, which is the case in your example.

If $A$ has equal eigenvalues $\lambda$, then $A$ and $A-\lambda I$ have same JCF, since $P^{-1}(A-\lambda I)P=P^{-1}AP-\lambda I$, thus your two matrices are equivalent to $A_1=\begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 \end{bmatrix}$ and $A_2=\begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\0 & 0 & 0 & 0 \end{bmatrix}$. Now you can see that they have same JCF, they both consists of one $2\times 2$ and two $1\times 1$ Jordan blocks. Also $\mathrm{dim}N(A_1^i)=\mathrm{dim}N(A_2^i)$ for $i=1,2,3,4$ and that $S^{-1}A_1S=A_2$, where permutation matrix $S=\begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \end{bmatrix}$.

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