# 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. – Arturo Magidin Dec 11 '19 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 $$S=\begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \end{bmatrix}$$.