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