Find characteristic polynomial and minimal polynomial of $T$.

Question

Let $V$ is a finite vector space, $T: V\rightarrow V$ is a linear operator with $KerT=ImT$. Find characteristic polynomial and minimal polynomial of $T$.

I just know that $dimV$ is even, I don't know how to use the condition$ImT=kerT$. Help me, thanks!

Answer 1

If $Ker T=Im T$, then $T^2=0$ so its minimal polynomial is $X^2$ if $T$ is not trivial. If the dimension of $V$ is $n$, the characteristic polynomial of $T$ is $X^n=0$ since $T$ is nilpotent.

Answer 2

For $v \in V$, $Tv \in \mathrm {Im} \ T = \ker T$, thus $T^2 v = 0$ for all $v \in V$. In other words, $T^2 = 0$. Thus the minimal polynomial $m(x)$ divides $x^2$.

If $m(x) = x$, then $T = 0$. But for the operator $0$, $\mathrm {Im} \ 0 = 0$, $\ker 0 = V$, contradiction. Thus $m (x) = x^2$.

The characteristic polynomial $f(x)$ has the same roots with the minimal polynomial, thus $$f(x) = 0 \iff x = 0. $$ Then $f(x) = x^{\dim V}$ since $f(x)$ is always a monic polynomial of degree $\dim V$.