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(a) Prove or give a counterexample: if $T\colon\mathbb{R}^3\to\mathbb{R}^3$ is a linear transformation such that $\mbox{null}(T)\cap\mbox{range}(T)$ has dimension at least $1$ then $T$ is nilpotent.

(b) Prove or give a counterexample: if $T\colon\mathbb{R}^4\to\mathbb{R}^4$ is a linear transformation such that $\mbox{null}(T)\cap\mbox{range}(T)$ has dimension at least $2$ then $T$ is nilpotent.

For part (a) consider a transformation $T(x,y,z)=(0,x,z)$. It's obviously linear, $\mbox{null}(T)=Y$ axis and $\mbox{range}(T)=Y\cup Z$ axes, however, it isn't nilpotent, since $T^3(x,y,z)=T^2(0,x,z)=T(0,0,z)=(0,0,z)$.

I believe it is also possible to come up with a counterexample to (b), but can't construct it.

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    $\begingroup$ Range is a span, not union of two subspaces. $\endgroup$
    – Marc van Leeuwen
    Jun 7, 2019 at 14:12
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    $\begingroup$ Hint: Apply the rank-nullity theorem. $\endgroup$
    – Bach
    Jun 7, 2019 at 14:40

1 Answer 1

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For (a), you should write $Y+Z$, not $Y\cup Z$. Apart from that, you are correct, $(0,0,1)$ is a fixed point of the operator so it can't be nilpotent.

Hint : For (b) what dimension must both $null(T)$ and $range(T)$ have for it to be possible ?

Then can you see why $null(T) = range(T)$, so $T^2=0$ ?

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  • $\begingroup$ Yes, $\mbox{null}(T)=\mbox{range}(T)$ follows from the rank-nullity theorem. $\endgroup$
    – Hasek
    Jun 7, 2019 at 14:50

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