For $n>2$, are there norms $\parallel.\parallel_{a}$ and $\parallel.\parallel_{b}$ on $M_{n}(\mathbb{R})$ with the following property:
$A\in M_{n}(\mathbb{R})$ is singular if and only if $\parallel A \parallel_{a}=\parallel A \parallel_{b}$
For $n=2,\;$ these norms are $\parallel A \parallel_{a}=\sqrt{\sum a_{ij}^{2}}$ and $\parallel A \parallel_{b}=\parallel.\parallel_{op}$, the operator norm.
(This answer expands the ideas in the comments by Mikael de la Salle and myself).
There is a family of matrix norms that can be defined as follows: given $p \geq 1,k \leq n$, $$ \|A\|_{p,k} = \left(\sum_{i=1}^k \sigma_i(A)^p\right)^{1/p}, $$ where we denote by $\sigma_i(A)$ the $i$th singular value of $A$ (taken in nonincreasing order, $\sigma_1(A)\geq \sigma_2(A)\geq \dots$). The most famous examples are the Schatten norms ($k=n$) and the Ky Fan norms ($p=1$), but they are matrix norms for each $k$ and $p$ (see e.g. this blog post by Nathaniel Johnston and the references included in it).
A $n\times n$ matrix $A$ is singular iff $\sigma_n(A)=0$, and this is equivalent to $\|A_{p,n}\|=\|A_{p,n-1}\|$ for every $p$. So any pair of norms with $k=n$ and $k=n-1$ works.