Let $V$ be a vector space of dimension $m\geq 2$ and $T: V\to V$ be a linear transformation such that $T^{n+1}=0$ and $T^{n}\neq 0$

Question

Let $V$ be a vector space of dimension $m\geq 2$ and $ T: V\to V$ be a linear transformation such that $T^{n+1}=0$ and $T^{n}\neq 0$ for some $n\geq1$ .Then choose the correct statement(s):

$(1)$ $rank(T^n)\leq nullity(T^n)$

$(2)$ $rank(T^n)\leq nullity(T^{n+1})$

Try:

I found this case is possible if $n<m$ and took some examples for $(2)$ , found it true but I've no idea how to prove. For (1) I'm not getting anything.

Answer 1

1.

Let $y\in Range(T)$ $\implies y=T(x)$ for some $x\in V$, $T^{n+1}(x)=T^{n}(T(x))=0 \implies y \in Ker T^n.$

2. Let $y \in KerT^n \implies T^n(y)=0 \implies T^{n+1}(y)=T(T^n(y))=T(0)=0 \implies y \in KerT^{n+1}$

Answer 2

$0=T^{n+1}(V)=T(T^n(V)) \implies rank(T^n)\le nullity (T)$. But of course $nullity T\le nullity T^{n+1}$

So (2) is true.

Oh and (1) is true also...

For (1), see here...

Answer 3

Note that $f=T^n$ is such that $f^2=0$. Thus $Im f \subseteq \ker f$ implies $rank(f) \le nullity(f) $, which is (1).