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.
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
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
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…