Every norm on M_n(R) is complete

Idea. Completeness is a property of the *uniform structure*, so it is the same for two equivalent norms. For one particular norm, the Frobenius norm, the space is just $\mathbb{R}^{n^2}$, which is complete.

Step 1: the Frobenius norm. A Cauchy sequence $(A_k)$ for $\|\cdot\|_F$ has Cauchy entries: $|a^{(k)}_{ij}-a^{(l)}_{ij}|\le\|A_k-A_l\|_F$ (previous problem). Each entry sequence converges in $\mathbb{R}$, say to $a_{ij}$. Let $A=(a_{ij})$; then $\|A_k-A\|_F^2=\sum_{i,j}(a^{(k)}_{ij}-a_{ij})^2\to0$, a finite sum of terms tending to $0$.

Step 2: any other norm. Let $\|\cdot\|'$ be any norm. By equivalence of norms there are $c,C>0$ with $c\|A\|_F\le\|A\|'\le C\|A\|_F$. A sequence is $\|\cdot\|'$-Cauchy iff it is $\|\cdot\|_F$-Cauchy, and it converges for one norm iff it converges for the other (to the same limit). Hence completeness transfers.

Remark. This is specific to finite dimension. In infinite dimension, different norms on the same vector space can give a complete and a non-complete space (e.g. continuous functions with the sup norm versus the $L^1$ norm).