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Let $A= \left\{ \begin{bmatrix} 0 & a_{12} & a_{13} \\ 0 & 0 & a_{23} \\ 0 & 0 & 0 \\ \end{bmatrix} \bigg|a_{12},a_{13},a_{23} \in \mathbb{C} \right\}$. Now define $\|A\|=\max_{k}\sum_{j=1}^{3} |a_{jk}|$ for all $a_{jk} \in \mathbb{C}$. It it straightforward to check that $A$ is normed space under the norm. But how to prove if it is Banach space? (i.e. complete under the norm).

I ask this question because not knowing where to start. So please give me a hint at least. Thank you!

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    $\begingroup$ $A$ is a finite-dimensional normed space over $\mathbb{C}$... $\endgroup$ – MSobak Sep 12 '18 at 10:21
  • $\begingroup$ @Sobi Thank you for the source! But can i prove it without using it? Is it hard? $\endgroup$ – Reza Habibi Sep 12 '18 at 10:28
  • $\begingroup$ For these kinds of things, I think it's actually simpler to prove the general result. That being said, it is of course possible to prove it without invoking the general result. $\endgroup$ – MSobak Sep 12 '18 at 10:29
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As I pointed out in the comments, this follows directly since you are dealing with a finite-dimensional normed space over a complete field (namely $\mathbb{C}$). However, if you want to prove it directly, you do it as follows.

Let $\{A^{(k)}\}_{k=1}^\infty$ be Cauchy, i.e. $$ \Vert A^{(n)} - A^{(m)} \Vert < \epsilon $$ when $n,m \geq N$ for some $N$. This is equivalent to saying $$ |a^{(n)}_{jk}-a^{(m)}_{jk}| \leq \max_k \sum_{j=1}^3 |a^{(n)}_{jk}-a^{(m)}_{jk}| < \epsilon, $$ whenever $n,m > N$, so that $\{a^{(n)}_{jk}\}_{n=1}^\infty$ is Cauchy in $\mathbb{C}$ for each $j,k$, and since $\mathbb{C}$ is complete, it converges, i.e. $a^{(n)}_{jk}\to \tilde a_{jk} \in \mathbb{C}$ as $n\to \infty$. Put $$ \tilde A = \begin{pmatrix}0 & \tilde a_{12} & \tilde a_{13}\\0 & 0 & \tilde a_{23}\\ 0&0&0\end{pmatrix}. $$ Then $\tilde A$ clearly belongs to your space. Can you now prove that $A^{(n)} \to \tilde A$ as $n \to \infty$ to finish the proof?

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  • $\begingroup$ It is clearly enough, really thank you. It helps me a lot! $\endgroup$ – Reza Habibi Sep 12 '18 at 10:40

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