Finding Jordan Canonical form given the minimal and characteristic polynomial. I have the following information: the characteristic polynomial of $A$ is $p_A(t)=(t-4)^3(t+6)^2$ and the minimal polynomial is $q_A(t)=(t-4)^2(t+6).$
I'm having problems seeing how one would work backwards with just this information.
My attempt at JCF.
$\begin{pmatrix}
4&1&0&0&0\\
0&4&0&0&0\\
0&0&4&0&0\\
0&0&0&-6&0\\
0&0&0&0&-6\\
\end{pmatrix}$
 A: The multiplicity of an eigenvalue as a root of the characteristic polynomial is the size of the block with that eigenvalue in the Jordan form. The size of the largest sub-block (Elementary Jordan Block) is the multiplicity of that eigenvalue as a root of the minimal polynomial.
Can you see how this determines the matrix?
Added: For example, the block with the eigenvalue $4$ in its diagonal must be $3\times3$, since that is the multiplicity of $3$ as a root of $p_A$. We also know by the definition of the Jordan form that this block is made up by smaller sub-blocks, each with $4$ in its diagonal and $1$s above, and with decreasing size as one goes down left to right. 
Now, without any further information, since the larger block is of size $3\times 3$, there are three possibilities for our subblocks:


*

*Three sub-blocks of $1 \times 1$, giving the following large block: $\left (\begin{matrix}4&0&0 \\0&4&0 \\0&0&4\end{matrix} \right )$

*One subblock of size two, and since the other must not be bigger and their sizes must "add up" to $3 \times 3$ we get: $\left (\begin{matrix}4&1&0 \\0&4&0 \\0&0&4\end{matrix} \right )$.

*A single large block: $\left (\begin{matrix}4&1&0 \\0&4&1 \\0&0&4\end{matrix} \right )$
But we know that the dimension of the largest sub-block must be the multiplicty of $4$ as a root of the minimal polynomial, hence the only possibility is: $$  \left (\begin{matrix}4&1&0 \\0&4&0 \\0&0&4\end{matrix} \right )$$
