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**NonCommAlg** see that $\displaystyle Ab=e_{n-2}, \ A^2b=Ae_{n-2}=e_{n-3}, \cdots, A^{n-2}b=e_1, \ A^{n-1}b=A^nb= \cdots = 0.$ now if $\displaystyle x \in \text{span} \{b,Ab,A^2b, \cdots, A^{n-2}b \},$ then:

$\displaystyle x=c_1e_1 + c_2e_2 + \cdots + c_{n-1}e_{n-1}.$ but then $\displaystyle Ax=\begin{cases} 0 & \text{if} \ \ n=2 \\ c_2e_1 + c_3e_2 + \cdots + c_{n-1}e_{n-2} & \text{if} \ \ n > 2 \end{cases},$ and obviously in either case $\displaystyle Ax \neq e_{n-1}=b.$ Q.E.D.