Show that the vector space $M_{nn}(\mathbb{R})$ which contains $n \times n$ Matrices is a Banach space.
Let be $\Vert \cdot\Vert_{max}= max_{1\leq i,j \leq n}\vert a_{ij}\vert$ the maximum of the absolute value of all entries of a matrix $A$.
My proof:
We show that every Cauchy sequence has a limit in $M_{nn}(\mathbb{R})$. I will construct such a limit matrix $C$ as follows:
Let $(A_m)_{m \to \infty}$ be a Cauchy sequence in $M_{nn}(\mathbb{R})$. So there exists a $m_0$ such that for every $m>m_0$ it holds: $\Vert A_m – A_{m_0}\Vert_{max} < \epsilon$, where $0<\epsilon$.
We define a sequence $(f_{m,i,j})_{m \to \infty}$ in $\mathbb{R}$ as follows:
$f_{m,i,j}= a_{ij}$, where $a_{ji}$ is the ij-th entry of Matrix $A_m$ which is the $m$-th member of the sequence $(A_m)_{m \to \infty}$.
Let be $f_{m_0,i,j}= b_{ji}$, where $b_{ji}$ is the $ij$-th entry of Matrix $A_{m_0}$. We then have for $f_m$ which consists of an arbitrary entry $a_{ij}$ of $A_m$: $\vert f_{m,i,j} – f_{m_0,i,j}\vert \leq max_{1\leq i, j \leq n}\vert a_{ij} – b_{ij} \vert = \Vert A_m – A_{m_0}\Vert_{max} < \epsilon $. So every sequence $f_m$ which is constructed from entries of $A_m$ is a Cauchy sequence and as we know that $\mathbb{R}$ is complete it is also convergent. Let $c$ be the limit of a sequence $f_{m,i,j}$. Let $C$ be a matrix in $M_{nn}(\mathbb{R})$which consists of all those $c$'s.
By definition of convergence there exists an index $m_0$ such that for a sequence $f_{m,i,j}$ which is constructed from a arbitrary $a_{ij}$ entry of $A_m$ we have: $\forall m > m_0: \vert f_{m,i,j} – c_{ij}\vert< \epsilon$. We then choose the greatest index $m_0$ out of all sequences $f_m$. We then consider our inital sequence $(A_m)_{m \to \infty}$. If we set our index: $m>m_0$ it yields $\Vert A_m – C \Vert_{max}= max_{1\leq i,j \leq n}\vert f_{m,i,j} – c_{ij}\vert< \epsilon$. So our sequence $(A_m)_{m \to \infty}$ is convergent which shows that every Cauchy sequence in $M_{nn}(\mathbb{R})$ is convergent and hence $M_{nn}(\mathbb{R})$ is a Banach space. q.e.d.
Although it is not very elegant and somewhat brute force I want to now if this serves as a rigorous proof or are there any flaws?
Any comments are welcome!
Best Answer
Globally, it is correct, but it could be improved: