Prove one situation happens Let $f\in End(V)$ of rank 2 where $V$ is over $\mathbb{C}^n$ field. Prove one of situation happens:
1) $f$ is diagonalised
2 $tr(f)$ is an eigenvalue of f
3) $\frac{1}{2} tr(f)$ is an eigenvalue of f 
where tr is trace
My answer: (I'm using the fact that $f$ is similar to matrix in Jordan form since it's over $\mathbb{C^n}$
we have three options regardless to eigenvalues:
1) we have $\lambda_1=a, \lambda_2=b$ and the rest eigenvalues are $0$ then $f$ is similar to diagonal matrix so f is diagonalized
2) we have $\lambda_1=\lambda_2=a$ and rest is $0$ and here we have
 two options $A$ will be similar to diagonal matrix or it would have one jordan block $2x2$ for $a$ but then we have $\frac{1}{2} tr f=a$
3) all eigenvalues will be $0$ then 2 cases either A is similar to Jordan matrix with blocks (one bock)$3x3$ and rest 0 or to (two blocks)$2x2$  and rest 0. in both cases we have $trf=0= eigenvalue$
 A: The fact that the rank is $2$ gives you that $0$ is an eigenvalue (if $n>2$) with geometric multiplicity$~n-2$. Therefore its algebraic multiplicity is at least $n-2$, so $X^{n-2}$ divides the characteristic polynomial$~\chi$; let the remaining factors be $X-a$ and $X-b$.The trace $\def\tr{\operatorname{tr}}\tr f$ is the sum of the roots of$~\chi$, so  $\tr f=a+b$.
Now $f$ will be diagonalisable (not "diagonalised") if all geometric multiplicities are equal to the algebraic multiplicities of the same eigenvalue. One has the following possibilities


*

*$0,a,b$ are all distinct; now the algebraic multiplicities of $a,b$ are $1$, and $f$ is diagonalisable.

*$0=a$ or $0=b$; by symmetry we may assume $0=a$, in which case $b=\tr f$ is an eigenvalue of$~f$.

*$0\neq a=b$. Now $a=b=\frac12\tr f$ is a (multiple) eigenvalue of $f$.


Note that in$~$2. both $b\neq0$ and $b=0$ are possible; either way $f$ will not be diagonalisable. By contrast in$~$3 it is possible (though not very likely) that $f$ is diagonalisable after all.
