Exercise 8.A.21

Proof. Let $W={𝔽}^{\infty }\times {𝔽}^{\infty }$ and define by

that is, $T$ applies the backward shift operator (call it $B$) on the first slot and forward shift operator (call it $F$) on the second slot. Thus, for each positive integer $k$, we have

and

Moreover $\mathrm{range}<!--FUNCTION\; APPLICATION-->B^k={𝔽}^{\infty }$ and $\mathrm{null}<!--FUNCTION\; APPLICATION-->F^k=\{0\}$. Note that $\mathrm{null}<!--FUNCTION\; APPLICATION-->B^k⊊\mathrm{null}<!--FUNCTION\; APPLICATION-->B^{k+1}$ and $\mathrm{range}<!--FUNCTION\; APPLICATION-->F^k⊋\mathrm{range}<!--FUNCTION\; APPLICATION-->F^{k+1}$. Thus

and

Hence $\mathrm{null}<!--FUNCTION\; APPLICATION-->T^k⊊\mathrm{null}<!--FUNCTION\; APPLICATION-->T^{k+1}$ and $\mathrm{range}<!--FUNCTION\; APPLICATION-->T^k⊋\mathrm{range}<!--FUNCTION\; APPLICATION-->T^{k+1}$. □