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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?

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  • $\begingroup$ 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. $\endgroup$ – user1551 Apr 17 '14 at 0:12
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There are six cases for the characteristic polynomial $\;p(x)\;$ and for the minimal one $\;m(x)\;$:

$$\begin{align*}\bullet&p(x)=(x-a)(x-b)(x-c)=m(x)\;,\;a,b,c\;\;\text {different. In this case the matrices are diagonalizable:}\\ \begin{pmatrix}a&0&0\\0&b&0\\0&0&c\end{pmatrix}\\{}\\ \bullet&p(x)=(x-a)^2(x-b)=m(x)\;,\;\;a\neq b. \;\text{In this case, the JCF for both}\\ \text{ matrices is }\\{}\\\begin{pmatrix}a&1&0\\0&a&0\\0&0&b\end{pmatrix}\\{}\\ \bullet&p(x)=(x-a)^2(x-b)\;,\;\;m(x)=(x-a)(x-b)\;,\;\;a\neq b. \text{ Here, the JCF}\\ \text{in both cases is}\\{}\\\begin{pmatrix}a&0&0\\0&a&0\\0&0&b\end{pmatrix}\\{}\\ \bullet&p(x)=(x-a)^3...\text{Check the three cases for}\;\;m(x)\end{align*}$$

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  • $\begingroup$ The last matrix should have a letter b in it, I think. $\endgroup$ – Ned Apr 14 '14 at 22:58
  • $\begingroup$ Good catch @Ned, thanks. $\endgroup$ – DonAntonio Apr 14 '14 at 23:17
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For each eigenvalue, its multiplicity as root of the characteristic polynomial gives the sum of the sizes of the Jordan blocks; its multiplicity as root of the minimal polynomial gives the size of the largest Jordan block.

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).

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