Finding minimal, characteristic polynomials and rational, jordan forms from the dimensions of nullspaces Suppose $A$ is an 8x8 matrix with entries over $\mathbb{C}$ such that 


*

*$\dim \ker(A-2I)=2$

*$\dim \ker(A-2I)^2=3$

*$\dim \ker(A-3I)=2$

*$\dim \ker(A-3I)^2=4$

*$\dim \ker(A-3I)^3=5$ 


Problem - What is the characteristic polynomial of $A$, the minimal polynomial of $A$, the Jordan form and the rational form of $A$? 
From (2), (5), the eigenvalues are $2,3$ of multiplicity 3,5. So the minimal polynomial $m(x)$ is $(x-2)^3(x-3)^5$ and must necessarily be the characteristic polynomial? How can I determine if there are 1s on the upper subdiagonal? I would like to see how (1), (3), (4) useful. 
 A: From the nested kernels:
$$\ker(A-\lambda I)\subset\ker(A-\lambda I)^2\subset\dots\subset\ker(A-\lambda I)^r=\ker(A-\lambda I)^{r+1}=\dots $$
if you set $\;d_i=\dim\ker(A-\lambda I)^i$, then $d_i-d_{i-1}\enspace (i\ge 1$ is equal to the number of Jordan blocks$J_\lambda$ of size $\ge i$.
In particular, $d_1$ is the number of Jordan blocks.
So in the present case, we have $2$ Jordan blocks $J_2$ and  $2$ Jordan blocks $J_3$. Furthermore there is 


*

*$3-2=1$ Jordan block $J_2$ of size $2$,

*$4-2=2$ Jordan blocks $J_3$ of size $\ge 2$, amongst which $5-4=1$ Jordan block of size $3.


Conclusion: The Jordan canonical for of the matrix is:

From this we see the minimal and characteristic polynomials of $A$ are, respectively:
$$(x-2)^2(x-3)^3, \qquad (x-2)^3(x-3)^5.$$
To compute the Frobenius normal form we need the similarity invariants of the matrix. These are polynomials $P_1, \dots, P_r\enspace(r\le 8)$ such that


*

*$P_{i+1}\mid P_i$ for all $i<r$

*$P_1$ is the minimal polynomial of $A$, $(x-2)^2(x-3)$

*$P_1\dotsm P_r=\chi_A=(x-2)^3(x-3)^5$


Thus either we have only two similarity invariants: $P_1$ and $P_2=(x-2)(x-3)^2$ or three: $P_1$, $P_2=(x-2)(x-3)$, $P_3=x-3$. Let's set $P_3=1$ in the first case.
We know $P_2P_3$ is the g.c.d. of the minors of $A-xI$ of order $8-2+1=7$ and $P_3$ is the g.c.d. of the minors of order $8-3+1=6$ of the same matrix. The difference between both cases lies in the minors of order $6$ being coprime or not…
