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!

  • 5
    $\begingroup$ $A$ 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

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?

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.