Finding characteristic and minimal polynomials and the Jordan normal form of $f$, knowing some relations for $f$.

Given a vector space $$V$$ of dimension $$4$$ and a base $$\{v_1,v_2,v_3,v_4\}$$, let $$f$$ be an endomorphism of $$V$$ such that $$f^3=0$$ and moreover $$f(v_1)=f(v_2)=v_3$$, $$f(v_3)=kv_4$$, and $$f(v_4)\in\left$$, where $$k$$ is a real parameter.

I should find the characteristic polynomial $$\chi_f$$, as well as the minimal one $$m_f$$, and the Jordan normal form of $$f$$, depending on $$k$$ and $$f(v_4)$$.

Now, here's my thoughts: $$f^3=0$$ means that either $$m_f(t)=t^2$$ or $$m_f=t^3$$; furthermore, $$0=f^3(v_1)=f^3(v_2)=f^2(v_3)=f(kv_4)=k\cdot f(v_4)$$so I guess one should consider separately $$k=0$$, which lets $$f(v_4)$$ be a parameter, and $$k\ne0$$ which forces $$f(v_4)=0$$.

How should I continue?

Note that $$f$$ cannot be a Jordan block can by of size 4, since $$f^3=0$$. Also $$f^3=0$$ forces all eigenvalues of $$f$$ to be zero, so the characteristic polynomial is $$\chi_f(\lambda)=\lambda^4$$.
Assume first that $$k\ne0$$. As you say, $$f(v_4)=0$$. Then $$v_1,v_3,v_4$$ form a basis for a $$3\times 3$$ Jordan block. So the Jordan form of $$f$$ is $$\begin{bmatrix} 0&1&0&0\\ 0&0&1&0\\ 0&0&0&0\\ 0&0&0&0\end{bmatrix}.$$ Take now the case $$k=0$$. If $$b\ne0$$, then $$x=\tfrac ab\,v_1+\tfrac a{b^2}\,v_3+v_4$$ satisfies $$f(x)=bx$$, so $$b$$ is an eigenvalue. But this is not possible, as it contradicts $$f^3=0$$. Thus $$b=0$$, and so $$f(v_4)=av_1$$ for some $$a$$. So, if $$a\ne0$$, then $$v_3,v_1,\tfrac1a\,v_4$$ is a basis for a $$3\times 3$$ Jordan block. If $$a=0$$, then $$f(v_4)=0$$. Then $$v_1,v_3$$ give you a $$2\times 2$$ Jordan block, and you can complete the basis with $$v_4$$ and $$v_1-v_2$$, both in the kernel of $$f$$.
In summary: the deciding factors are $$k$$ and $$a$$ in $$f(v_4)=av_1+bv_4$$.
• If $$k=0$$ and $$a=0$$, the Jordan form is $$\begin{bmatrix} 0&1&0&0\\ 0&0&0&0\\ 0&0&0&0\\ 0&0&0&0\end{bmatrix}.$$
• If $$k=0$$ and $$a\ne0$$, or if $$k\ne0$$, the Jordan form is $$\begin{bmatrix} 0&1&0&0\\ 0&0&1&0\\ 0&0&0&0\\ 0&0&0&0\end{bmatrix}.$$