[Math] A Linear Operator of Rank 1

Question

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?

Answer 1

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 $T\ne0$.