# $3 \times 3$ matrices completely determined by their characteristic and minimal polynomials

How do you show that two $3 \times 3$ matrices with the same characteristic and minimal polynomials both conjugate to the same Jordan normal form, assuming no knowledge of the eigenspaces?

I know that it is possible to determine completely the Jordan normal form of a matrix only with its minimal and characteristic polynomial, up to dimension $6$, but only if one can compute the dimension of the eigenspace as well.

And why does this characterization fail for $4 \times 4$ matrices?

• If one more piece of information is given, namely, the dimension of eigenspace for each eigenvalue, the characterisation works up to 6x6 matrices. See Dennis Gulko's answer in another thread. Apr 17, 2014 at 0:12

There are six cases for the characteristic polynomial $\;p(x)\;$ and for the minimal one $\;m(x)\;$:
\begin{align*}\bullet&p(x)=(x-a)(x-b)(x-c)=m(x)\;,\;a,b,c\;\;\text {different. In this case the matrices are diagonalizable:}\\ \begin{pmatrix}a&0&0\\0&b&0\\0&0&c\end{pmatrix}\\{}\\ \bullet&p(x)=(x-a)^2(x-b)=m(x)\;,\;\;a\neq b. \;\text{In this case, the JCF for both}\\ \text{ matrices is }\\{}\\\begin{pmatrix}a&1&0\\0&a&0\\0&0&b\end{pmatrix}\\{}\\ \bullet&p(x)=(x-a)^2(x-b)\;,\;\;m(x)=(x-a)(x-b)\;,\;\;a\neq b. \text{ Here, the JCF}\\ \text{in both cases is}\\{}\\\begin{pmatrix}a&0&0\\0&a&0\\0&0&b\end{pmatrix}\\{}\\ \bullet&p(x)=(x-a)^3...\text{Check the three cases for}\;\;m(x)\end{align*}
Since the sizes of the Jordan blocks define a partition of their sum, the question is for which values of $n\geq m$ there is a unique partition of $n$ with largest part$~m$. This is true for all $m\leq n\leq3$, the partitions of$~3$ being $(3),(2,1),(1,1)$, all with different largest part. It fails first for $n=4,m=2$, which allows the two partitions $(2,2)$ and $(2,1,1)$. So an eigenvalue with multiplicity $4$ in the characteristic polynomial and multiplicity $2$ in the minimal polynomial would not allow determining the sizes of all Jordan blocks. This can happen with square matrices of size at least$~4$ (for some pair of characteristic and minimal polynomials).