Problem
Let $V$ be a three-dimensional vector space over $\mathbb{C}$ and let $f,g \in End_{\mathbb{C}}(V)$ with $f^3 = g^3 = 0$. In addition, let $\mu_f$ and $\mu_g$ be the minimal polynomial of $f$ and $g$ respectively. Proof that $\mu_f = \mu_g$ if and only if there exists some $h \in End_{\mathbb(C)}(V)$ so that $h^{-1}\cdot f \cdot h=g$.
My Question
Hello, StackExchange, I think I solved this excersize at least half decently, but I know something must be wrong, because my proof doesn't use the condition $f^3 = g^3 = 0$. Could you give me a hint where I made a mistake? Thank you very much.
My Proof
"$\Rightarrow$"
We have $\mu_f = \mu_g$. Since we are in a three-dimensional vector space, we can look at the possible Jordan Normal Forms of $f$ and $g$.
Possibility 1
$deg(\mu_f)=deg(\mu_g)=1$
In this case, $f$ and $g$ are diagonalizable and have the share the same diagonal matrix. Therefore, they are similar which is the statement we wanted to show.
Possibility 2
$deg(\mu_f)=deg(\mu_g)=2$ and $f,g$ have two Eigenvalues.
In this case, $f,g$ are again diagonalizable.
Possibility 3
$deg(\mu_f)=deg(\mu_g)=2$ and $f,g$ have only one Eigenvalue.
In this case, $f,g$ are not diagonalizable, but they share the same Jordan Normal Form and again they are therefore, similar.
Possibility 4-6
Since my argumentation will stay the same, I won't bother you with it.
"$\Leftarrow$"
We have
$$H^{-1} \cdot A_f \cdot H = B_g$$.
Since $f$ is an endomorphism, it must have a Jordan Normal Form such as:
$$A_f = X \cdot J_{A_f} \cdot X^{-1}$$
$$\Rightarrow H^{-1} \cdot X \cdot J_{A_f} \cdot X^{-1} \cdot H = B$$
Now, we can verify that $(H^{-1} \cdot X)^{-1} = X^{-1} \cdot H$. Therefore, $A_f$ and $B_g$ share the same Jordan Normal Form. And again, since we are in a three-dimensional vector space, the minimal polynomial of $f$ and $g$ must be the same (I would again argue with checking all 6 possible Jordan Normal Forms).
If $f,g$ are not assumed nilpotent, then $$f=\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\end{pmatrix},\qquad g=\begin{pmatrix}1&0&0\\0&1&0\\0&0&0\end{pmatrix}$$ have $\mu_f=\mu_g=X^2-X$. As you say, they are both diagonalizable - but they are not similar.