I want to ask a question regarding classes of similar $3 \times 3$ matrices. We were told that two matrices are similar if and only if they have the same Jordan normal form. That led me to trying to classify all endomorphisms in $\mathbb{R}^3$ up to similarity (or choice of base) using their Jordan normal form. I've arrived at the following table.
| $\chi_A$ | $\mu_A$ | partition | jordan |
|---|---|---|---|
| $$(X-a)^3$$ | $$(X-a)$$ | $$(1,1,1)$$ | $$Diag(a,a,a)$$ |
| $$(X-a)^2$$ | $$(2,1)$$ | $$Diag(J_2(X-a),a)$$ | |
| $$(X-a)^3$$ | $$(3)$$ | $$Diag(J_3(X-a))$$ | |
| $$(X-a)^2\cdot (X-b)$$ | $$(X-a)\cdot (X-b)$$ | $$(1,1)(1)$$ | $$Diag(a,a,b)$$ |
| $$(X-a)^2\cdot (X-b)$$ | $$(2)(1)$$ | $$Diag(J_2(X-a),J_1(X-b)$$ | |
| $$(X-a)\cdot (X-b) \cdot (X-c)$$ | $$(X-a)\cdot (X-b)\cdot (X-c)$$ | $$(1)(1)(1)$$ | $$Diag(a,b,c)$$ |
| $$(X-a)\cdot (X^2+bX+c)$$ | $$(X-a)\cdot (X^2+bX+c)$$ | $$(1),(1)$$ | $$Diag(J_1(X-a),J_1(X^2+bX+c))$$ |
Now I'm a bit confused because I arrived at the result, that for each pair of characteric and minimal polynomial there is only exactly one jordan normal form, so all $3 \times 3$ matrixes $A, B$ with $\mu_A = \mu_B$ and $\chi_A = \chi_B$ would be similar (and vice versa), which doesn't hold generally. Is this table correct and does that result hold? I falsely asked the same (a similar) question on MathOverflow already and was pointed to the rational canonical form (RCF). The RCF doesn't change any of my results however, right? Matrices with the same Jordan normal form should have the same RCF if I understood the RCF correctly.