# 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.

$\begin{pmatrix} 4&1&0&0&0\\ 0&4&0&0&0\\ 0&0&4&0&0\\ 0&0&0&-6&0\\ 0&0&0&0&-6\\ \end{pmatrix}$

• NB in general the minimal and characteristic polynomials of a matrix do not determine its Jordan form uniquely. For example, there are two Jordan matrices $J$ (up to rearrangement) for which $p_A(t) = t^4$ and $m_A(t) = t^2$. – Travis Willse Jul 22 at 2:44

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: $$\left (\begin{matrix}4&1&0 \\0&4&0 \\0&0&4\end{matrix} \right )$$

• I have a matrix that is the eigenvalues on the diagonal, 1's above the eigenvalues, and zeroes everywhere else. I sort of see this relationship. – emka Mar 31 '15 at 2:30
• I added to my original answer how to figure out the form of the block with $4$s in its diagonal. – Reveillark Mar 31 '15 at 2:38
• I put an attempt above. – emka Mar 31 '15 at 2:46
• Your attempt is correct :) I'm assuming you used that the multiplicty of $-6$ as a root of the minimal is $1$ and hence there are two sub-blocks of sizes $1\times 1$ each. – Reveillark Mar 31 '15 at 2:48
• I was very tempted to stick a 1 in the lower part of that matrix. It took me a rather long moment to make the connection. Your prompts were very helpful. – emka Mar 31 '15 at 2:50