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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.

  • I found this case is possible if $n<m$ can you explain why –  Jan 24 '18 at 12:43
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    @ManeeshNarayanan If $T^k=0$ for some $k$, then already $T^m=0$. It's a general fact about nilpotent endomorphisms of $m$-dimensional spaces. Thus, if $n\ge m$, it's impossible that $T^n\ne0$ and $T^{n+1}=0$. – egreg Jan 24 '18 at 13:12

3 Answers3

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

So (2) is true.

Oh and (1) is true also...

For (1), see here...

  • what about $(1)$.. –  Jan 24 '18 at 08:26
  • Certainly $0 = T^{n+1}(V)$ implies the null space of $T^{n+1}$ is the full space $V$, so $nullity(T^{n+1}) =dim(V)$ which implies the conclusion that (2) is true. However, I cannot see how your alternative reasoning works: What relevance does $T(T^n(V))$ have in your argument, and how do you arrive at the "implies" sign? – Michael Jan 24 '18 at 09:20
  • @Michael the image of $T^n$, whose dimension is $rank(T^n)$, is a subspace of the null space of $T^{n+1}$... –  Jan 24 '18 at 09:28
  • The null space of $T^{n+1}$ is everything. So what is the business of $T(T^n(V))$? [In other words, your construction seems to be $2+2=4$ so $d/dx \sin(x) = \cos(x)$. Both are true but I cannot follow the logical reasoning.] – Michael Jan 24 '18 at 09:32
  • @Michael $T(T^n(V))=0$ says precisely that $T^n(V)\subset kerT$... I see now that I skipped a step... namely $kerT\subset kerT^2\subset \cdots kerT^{n+1}$... –  Jan 24 '18 at 09:57
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    I see your reasoning now, your last comment explains your original reasoning behind why you used $T(T^n(V))$. Though it does seem a bit of overkill, since the null space of $T^{n+1}$ is the entire space so we are done (no subspaces can have larger dimension than the whole space). – Michael Jan 24 '18 at 10:25
  • (2) is obviously true. Your argument, with $\operatorname{nullity}T^n$ at the end proves (1). – egreg Jan 24 '18 at 13:11
  • @egreg good insight... –  Jan 24 '18 at 14:20
  • @Michael good point –  Jan 24 '18 at 14:21
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1.

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

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

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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).

Andrea Marino
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