How many Jordan normal forms are there when the characteristic polynomial is $(\lambda+4)^5(\lambda-2)^2$?

Question

Let $A\in M_7(\mathbb{C})$ be a matrix in with the characteristic polynomial $p(A)=(\lambda+4)^5(\lambda-2)^2$.

I need to find all Jordan normal forms for this.

I think that i can use that the Jordan normal form of a given matrix $A$ is unique up to the order of the Jordan blocks, so because of that I think that there are 14 but i only find 10 so i'm almost sure that something is wrong.

I think that there are $14$ because if i undertood the above theorem there are $12$ for the $(\lambda+4)^5$ and $2$ for $(\lambda-2)^2$.

Answer 1

The 14 forms are

$m(A)=(\lambda+4)^5(\lambda-2)^2$

$m(A)=(\lambda+4)^4(\lambda-2)^2$

$m(A)=(\lambda+4)^3(\lambda-2)^2$ with decomposition 3,2

$m(A)=(\lambda+4)^3(\lambda-2)^2$ with decomposition 3,1,1

$m(A)=(\lambda+4)^2(\lambda-2)^2$ with decomposition 2,2,1

$m(A)=(\lambda+4)^2(\lambda-2)^2$ with decomposition 2,1,1,1

$m(A)=(\lambda+4)(\lambda-2)^2$

$m(A)=(\lambda+4)^5(\lambda-2)$

$m(A)=(\lambda+4)^4(\lambda-2)$

$m(A)=(\lambda+4)^3(\lambda-2)$ with decomposition 3,2

$m(A)=(\lambda+4)^3(\lambda-2)$ with decomposition 3,1,1

$m(A)=(\lambda+4)^2(\lambda-2)$ with decomposition 2,2,1

$m(A)=(\lambda+4)^2(\lambda-2)$ with decomposition 2,1,1,1

$m(A)=(\lambda+4)(\lambda-2)$