Find Jordan form from characteristic and minimal polynomial

Question

I know the characteristic polynomial of $A$ is $k_A(t)=t^6(t-1)^4(t-2)^2$

and the minimal polynomial is $m_A(t)=t^5(t-1)(t-2).$

I dont really know what to do.

I tried this:

In the Jordan form there must be six $0$-s, four $1$-s and two $2$-s in the diagonal, because of the characteristic polynomial.

If i consider the $0$-s: the size of the block is $5$, because of the minimal polynomial?

Same with $1$-s and $2$-s, so i get:

$\begin{pmatrix} 2&0&0&0&0&0&0&0&0&0&0&0 \\ 0&2&0&0&0&0&0&0&0&0&0&0 \\ 0&0&1&0&0&0&0&0&0&0&0&0 \\ 0&0&0&1&0&0&0&0&0&0&0&0 \\ 0&0&0&0&1&0&0&0&0&0&0&0 \\ 0&0&0&0&0&1&0&0&0&0&0&0 \\ 0&0&0&0&0&0&0&1&0&0&0&0 \\ 0&0&0&0&0&0&0&0&1&0&0&0 \\ 0&0&0&0&0&0&0&0&0&1&0&0 \\ 0&0&0&0&0&0&0&0&0&0&1&0 \\ 0&0&0&0&0&0&0&0&0&0&0&1 \\ 0&0&0&0&0&0&0&0&0&0&0&0 \\ \end{pmatrix}$

Is this correct? I would be glad if someone could explain me in detail how to do it, and if im right why am i right?

I dont really understand this whole characteristic polynomial-minimal polynomial-jordan form stuff

Answer 1

First of all we can do some consideration regards the relation between the minimal polynomial $m_{A}(t)$ and characteristical polynomial $p_{A}(t)$ in general :

Of course for Theorem of Primary Decomposition we can restrict of our view (beeing $A$ invariants) to single generalized eigenspaces $V_{\lambda}' = Ker(f-\lambda I_{d})^{k}$

(Where $k$ denote the algebraic multiplicity of $\lambda$ in the characteristical polynomial, for each $\lambda \in Sp(f)$).

That beeing said, in each block if we know that the caracteristical and the minimal polynomial we can draw some conclusion, which in this are are even sufficient.

From the Jordan normal form construction, we know that the degree of the minimal polynomial corresponds to the higher size block in Jordan normal form (within each $V_{\lambda}$).

Studying separately our $V_{\lambda}'$,

We notice that $dim(V_{2}') = 2$ and the mimimal polynomial of $V_{2}'$ is $m_{A_{|_{V_{2}'}}}(t) =(t-2)$. As mention above the greater block must be of size $1$, which leads to the canonical Jordan form of $A_{|_{V_{2}'}}$ beeing $$\begin{pmatrix} 2 & 0 \\ 0 & 2\end{pmatrix}$$

Another way could be noticing that if the minimal polynomial factors completely (always speaking inside a general $V_{\lambda}'$) it must follow from this that the associated matrix is diagonalizable.

In this we can conlude for the cited theorem that being $m_{A_{|_{V_{1}'}}} = (t-1)$ the Jordan normal form of $A_{|_{V_{1}'}}$ is $$\begin{pmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{pmatrix}$$

In the end since the characteristical polynomial of $V_{0}' = t^{6}$ and $m_{A_{|_{V_{0}'}}} = t^{5}$ there must be a block of size $5\times5$ and just one (since dimension of $V_{0}'$ is 6) of size $1\times1$, which give us the Jordan normal form of $A_{|_{V_{0}'}}$ beeing $$\begin{pmatrix}0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0\end{pmatrix}$$

Since we want uniqueness (unless permutation of blocks) of Jordan normal form, we can arrange the block in increasing order to get $the$ Jordan normal form.