# Thread: finding characteristic polynomial, problem with the question?

1. ## finding characteristic polynomial, problem with the question?

Hi.
I was asked to find the characteristic polynomial of A.
A has the following attributes:

$A\in M_n(F)$

$rankA=1$

$n\geq 2$

$traceA=0$

I argue that there is a contradiction here!

if $rankA=1$, it means that: $dimV_0=dimKerA=n-1$.
if $dimKerA=n-1$ then it means that the diagonal matrix D (who has the eignvalues on it's main diagonal), has some $\lambda$ in its $d_{1,1}$ entry, and the rest are zeroes. meaning, D looks like this:
$\begin{bmatrix}\lambda & 0 & . & . & . & 0\\ 0 & 0 & & & & .\\ . & & 0 & & & .\\ . & & & 0 & & .\\ . & & & & 0 & .\\ 0 & 0 & . & . & . & 0\end{bmatrix}$
now, the problem is, that $traceA=0$, and since the trace is the sum of the eignvalues, we'll get that $\lambda=0$, and that means that $rankA=0$! (and not 1 as stated in the question)

2. ## Re: finding characteristic polynomial, problem with the question?

Hi,
I think you'd best rethink your analysis. Does the following matrix satisfy the conditions?

$A=\begin{bmatrix}1&1\\-1&-1\end{bmatrix}$

What is the characteristic polynomial of A?

3. ## Re: finding characteristic polynomial, problem with the question?

OK, now I see what was wrong with my analysis.
I thought that elementary row operations doesn't change the transformation.
dunno why...