Finding the characteristic polynomial of a linear transformation of rank $2$ Let $V$ be a finite dimensional vector space, $f \in L(V)$  with rank $2$. I need some help to prove that: $$\chi_f(x)=x^{n-2}\left(x^2-\textrm{tr}(f)x+\cfrac{\textrm{tr}(f)^2}{2}-\cfrac{1}{2}\textrm{tr}(f^2)\right).$$
For example, how can we use the rank $2$?
 A: Using rank-nullity theorem, one has:
$$\dim(\ker(f))=n-2.$$
Therefore, $0$ is an eigenvalue of $f$ of multiplicity $n-2$ and $x^{n-2}\vert\chi_f$ (the algebraic multiplicity is at least the geometric one). Recall that $\chi_f$ is a monic polynomial of degree $n$, hence:
$$\chi_f=x^{n-2}(x^2+\lambda x+\mu).$$
It is know that the coefficient of degree $n-1$ in $\chi_f$ is $-\textrm{tr}(f)$, expand $\det(x\textrm{id}-f)$, whence:
$$\chi_f=x^{n-2}(x^2-\textrm{tr}(f)x+\mu).$$
One has still to find the coefficient of $n-2$ in $\chi_f$.
Working in the splitting field of $\chi_f$, you can assume without loss of generality that $f$ is tridiagonal, then:
$$\mu=\sum_{i,j}f_{i,i}f_{j,j}.$$
Now notice that:
$$\textrm{tr}(f)^2-\textrm{tr}(f^2)=\left(\sum_{i}{f_{i,i}}\right)^2-\sum_{i}{f_{i,i}}^2=2\sum_{i,j}f_{i,i}f_{j,j}.$$
Whence the result.

If $V$ is a real vector space, you can see $f$ as a linear transformation of $V$ seen as a complex vector space, so that $\chi_f$ splits and $f$ is trigonalizable. Then, use the fact that similar endomorphisms have the same trace and characteristic polynomial. This is exactly what I have done for vector spaces over general fields. 
