The eigenvalues are $3$ (with multiplicity 5) and $2$ (with multiplicity 4).
The number of Jordan blocks associated to $3$ is equal to the dimension of the eigenspace of $3$, namely, to $\mathrm{nullity}(A-3I)$. By the Rank-Nullity Theorem, since $A$ is $9\times 9$ and the rank of $A-3I$ is $7$, the nullity is $2$, so there will be two Jordan blocks associated to $3$.
How many blocks are there of size at least 1? The nullity of $A-3I$. How many blocks are there of size at least 2? The initial vectors and second vectors of every block will be annihilated by $(A-3I)^2$; so you can figure out how many blocks have size at least $2$ by comparing the nullity of $A-3I$ and the nullity of $(A-3I)^2$. Every "extra" dimension you get in $(A-3I)^2$ gives you a block of size at least $2$. We have enough information to know that the nullity of $(A-3I)^2$ is 4; that means that there are 4 vectors in a Jordan canonical basis that are annihilated by $(A-3I)^2$. Two of them are the eigenvectors we already had, so that leaves $2$ vectors that correspond to blocks of size at least $2$. Continuing this way, we see that since the nullity of $(A-3I)^3$ is $5$, we get one more vector than before. That means that there is one block of size at least $3$ (four of the vectors were already accounted for, so you just add one more); the nullity of $(A-3I)^4$ is also $5$, so there are no new vectors added. There are no blocks of size at least $4$.
So we have two Jordan blocks associated with $3$: one block of size $2$, one block of size $3$.
Do something similar with $\lambda=2$.
This can all be visualized with the dot diagram of $A$ associated to $3$. This is an array of dots where there is a column for each block, and the number of rows in each column represents the length of the corresponding cycle.
If column $i$ has $p_i$ rows, then we will have $p_1\geq p_2\geq\cdots\geq p_k$ (where $k$ is the number of columns). Let $r_1\leq r_2\leq\cdots\leq r_{p_1}$ be the number of dots in each row. You can reconstruct the diagram on the basis of the $p_i$ or on the basis of the $r_j$. The following formulas give you the $r_i$:
$\begin{align*} r_1 &= \dim(V) - \mathrm{rank}(T-\lambda I)\\ r_j &= \mathrm{rank}((T-\lambda I)^{j-1}) - \mathrm{rank}((T-\lambda I)^j)\quad\text{if }j\gt 1. \end{align*}$ In your example, for $\lambda=3$, we would have $\begin{align*} r_1 &= 9-7 = 2\\ r_2 &= 7-5 = 2\\ r_3 &= 5-4 = 1\\ r_4 &= 4-4 = 0. \end{align*}$ So the dot diagram associated to $\lambda=3$ is: $\begin{array}{ccc} \bullet &\quad & \bullet\\ \bullet & \quad & \bullet\\ \bullet\\ \end{array}$ The first column tells you there is a block of length $3$; the second column that there is a block of length $2$; the lack of a third column tells you there are only two blocks.
The $\bullet$ correspond to vectors in Jordan canonical basis. If $v_1$ and $v_2$ are the end vectors of the Jordan cycles associated to the two blocks, the dots correspond as follows: $\begin{array}{lcl} \bullet\ (A-3 I)^{3-1}(v_1) &\quad&\bullet\ (A-3 I)^{2-1}(v_2)\\ \bullet\ (A-3 I)^{3-2}(v_1) &\quad&\bullet\ v_2\\ \bullet\ v_1 \end{array}$