# Is a linear transformation with non-empty kernel and image intersection necessarily nilpotent?

(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.

• Range is a span, not union of two subspaces. Jun 7, 2019 at 14:12
• Hint: Apply the rank-nullity theorem.
– Bach
Jun 7, 2019 at 14:40

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 $$T^2=0$$ ?
• Yes, $\mbox{null}(T)=\mbox{range}(T)$ follows from the rank-nullity theorem. Jun 7, 2019 at 14:50