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Given that $E$ is a finite dimensional space. Let $\dim E=n\geq 1$ and $\{e_i\}^{n}_{i=1}$ be a basis for $E.$ Then, there exists unique scalars $\{\alpha_i\}^{n}_{i=1}$ such that \begin{align}x=\sum_{i=1}^{n}\alpha_i e_i.\end{align} The problem is: I want to prove that $\| \cdot \|_0$ defined by $\| x \|_0=\max\limits_{1\leq i\leq n}|\alpha_i|$ is a norm on $E$.

I, therefore, post the proof in the answer section after it has been approved.

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  • $\begingroup$ Your problem makes no sense. What are the $\alpha_i$'s? $\endgroup$ – José Carlos Santos Dec 26 '18 at 11:58
  • $\begingroup$ @ José Carlos Santos: They are unique scalars such $\{e_i\}^{n}_{i=1}$ is a basis for $E$. $\endgroup$ – Omojola Micheal Dec 26 '18 at 12:00
  • $\begingroup$ That answer makes no sense either. $\endgroup$ – José Carlos Santos Dec 26 '18 at 12:01
  • $\begingroup$ Okay, let me edit my post then. Just some time! $\endgroup$ – Omojola Micheal Dec 26 '18 at 12:02
  • $\begingroup$ @José Carlos Santos: Kindly check, I made some edits. It should be fine now! $\endgroup$ – Omojola Micheal Dec 26 '18 at 12:07
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$a.\qquad$ Let $x\in E,$ then \begin{align}\|x\|_0=0&\iff \max\limits_{1\leq i\leq n}|\alpha_i|=0 \iff |\alpha_i|=0,\,\forall \,{1\leq i\leq n}\\&\iff \alpha_i=0,\,\forall \,{1\leq i\leq n}\\&\iff x=0,\;\forall\;x\in E\end{align} $b.\qquad$ Let $\lambda \in K$, then $\lambda x=\sum_{i=1}^{n}\lambda\, \alpha_i e_i$ and

\begin{align}\|\lambda x\|_0&= \max\limits_{1\leq i\leq n}|\lambda\,\alpha_i|=|\lambda\,|\max\limits_{1\leq i\leq n}|\alpha_i|=|\lambda\,|\| x\|_0,\;\forall\;x\in E\end{align}

$c.\qquad$ Let $x,y\in E$, then $x=\sum_{i=1}^{n}\alpha_i e_i$ and $y=\sum_{i=1}^{n}\beta_i e_i,$ for some $\alpha_i,\beta_i,\;i\in\{1,2,\cdots,n\}.$ Thus,

\begin{align}\| x+y\|_0&= \max\limits_{1\leq i\leq n}|\alpha_i+\beta_i|\leq \max\limits_{1\leq i\leq n}|\alpha_i|+\max\limits_{1\leq i\leq n}|\beta_i|=\| x\|_0+\|y\|_0,\;\forall\;x,y\in E\end{align}

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