# Possible Jordan Canonical Forms

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 $$(x-1)^{2}$$.

• 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. Dec 11, 2019 at 17:03

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 $$S=\begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \end{bmatrix}$$.