A Linear Operator of Rank 1 Let $T$ be a linear operator with rank $1$ on a finite dimensional vector space $V$.Then Which of the following are true?
1)either $T$ is diagonalizable or  $T$ is nilpotent.
2)$T$ is both diagonalizable and  nilpotent.

I take $T$ as constant mapping and got it as diagonalizable. So can we say that 1) is true and 2) is false? Is there any other method to solve the problem?

 A: We have $\dim\ker T=\dim V-1=n-1$ since the rank of $T$ is $1$ hence $0$ is an eigenvalue with multiplicity at least $n-1$ and then there are two possibilities:


*

*If $\operatorname{tr}(T)\ne 0$ then the last eigenvalue is $\lambda=\operatorname{tr}(T)$ and $T$ is diagonalizable since
$$V=\ker T\oplus E_\lambda(T)$$

*$\operatorname{tr}(T)= 0$ then $0$ is the only eigenvalue of $T$ so it's nilpotent and not diagonalizable since $T\ne0$.

A: 1) if $T$ is rank one, then $T=uv^T$ with nonzero $u,v$. If $v^Tu=0$ then $T$ is nilpotent. Otherwise $T$ is diagonalizable with eigenvalues $v^Tu$ (for eigenvector $u$) and eigenvalue $0$ (eigenvectors are all vectors orthogonal to $v$).
2) is definitely false as a diagonalizable and nilpotent mapping must be zero.
A: if $A$ is rank one, then $A = uv^\top$ for some $u, v \neq 0.$ the eigenvalues are $v^\top u, 0, 0, \cdots, 0$ and the corresponding  eigenvctors are $u$ and a basis of $v^\perp.$  since $range$ and kernel of $A$ fill up the space if $u$ and $v$ are not orthogonal, $A$ is diagonalizable. 
in the case where $u$ and $v$ are orthogonal, you have a nilpotent nondiagonalisable rank one matrix. for example, $u = \pmatrix{1\\0}, v = \pmatrix{0\\1}, uv^\top=\pmatrix{0&1\\0&0}.$ 
