Let $(e_{1}^{\vee},\ldots,e_{r}^{\vee})$ be the dual basis of $(e_{1},\ldots,e_{r})$. For $a_{1},\ldots,a_{r}\in k$ with $a_{i}\not=0$,

$$\frac{|(e_{i}^{\vee})(a_{1}e_{1}+\cdots+a_{r}e_{r})|}{\|a_{1}e_{1}+\cdots+a_{r}e_{r}\|}\leq\frac{|a_{i}|}{\alpha\max\{|a_{1}|\|e_{1}\|,\ldots,|a_{r}|\|e_{r}\|\}}\leq\frac{|a_{i}|}{\alpha|a_{i}|\|e_{i}\|}=\frac{1}{\alpha\|e_{i}\|},$$

and hence $\|e_{i}^{\vee}\|^{\vee}\leq(\alpha\|e_{i}\|)^{-1}$. Therefore, for $a^{\prime}_{1},\ldots,a^{\prime}_{r}\in k^{\prime}$,

	$\displaystyle\|a^{\prime}_{1}e_{1}+\cdots+a^{\prime}_{r}e_{r}\|$	$\displaystyle\geq\frac{|(e_{i}^{\vee}\otimes 1)(a^{\prime}_{1}e_{1}+\cdots+a^{\prime}_{r}e_{r})|^{\prime}}{\|e_{i}^{\vee}\|^{\vee}}$
		$\displaystyle=\frac{|a^{\prime}_{i}|^{\prime}}{\|e_{i}^{\vee}\|^{\vee}}\geq\frac{|a^{\prime}_{i}|^{\prime}}{(\alpha\|e_{i}\|)^{-1}}=\alpha|a^{\prime}_{i}|^{\prime}\|e_{i}\|.$

Thus we have the assertion. ∎