Hi.

I was asked to find the characteristic polynomial of A.

A has the following attributes:

$\displaystyle A\in M_n(F)$

$\displaystyle rankA=1$

$\displaystyle n\geq 2$

$\displaystyle traceA=0$

I argue that there is a contradiction here!

if $\displaystyle rankA=1$, it means that: $\displaystyle dimV_0=dimKerA=n-1$.

if $\displaystyle dimKerA=n-1$ then it means that the diagonal matrix D (who has the eignvalues on it's main diagonal), has some $\displaystyle \lambda$ in its $\displaystyle d_{1,1}$ entry, and the rest are zeroes. meaning, D looks like this:

$\displaystyle \begin{bmatrix}\lambda & 0 & . & . & . & 0\\ 0 & 0 & & & & .\\ . & & 0 & & & .\\ . & & & 0 & & .\\ . & & & & 0 & .\\ 0 & 0 & . & . & . & 0\end{bmatrix}$

now, the problem is, that $\displaystyle traceA=0$, and since the trace is the sum of the eignvalues, we'll get that $\displaystyle \lambda=0$, and that means that $\displaystyle rankA=0$! (and not 1 as stated in the question)