I'm interested in the maximal dimension of a subspace $V\leq\mathbb R^{n\times n}$ in which every nonzero matrix is invertible.
Odd $n$: For odd $n$ the maximum is $1$: if $A$ and $B$ would be linearly independent, then $\det(xA+B)$ is an odd polynomial of degree $\leq n$ with leading coefficient $\det A\neq0$, so it has a real root $x$; hence $0\neq xA+B$ is not invertible.
Even $n$: For $n$ even the maximum can be no larger than $n$: if $A_1,\ldots,A_{n+1}$ would be linearly independent, then the first row of every nontrivial linear combination must be nonzero, meaning that the first rows of $A_1,\ldots,A_{n+1}$ are linearly independent, contradicting $\dim\mathbb R^n=n$.
For $n=2$ the maximum is $2$, as $\begin{vmatrix}a&-b\\b&a\end{vmatrix}=a^2+b^2\neq0$ for $a,b$ not both zero.
For $n=4$ the maximum is $4$, as $$\begin{vmatrix}-a&b&c&d\\ b&a&-d&c\\ c&d&a&-b\\ d&-c&b&a\end{vmatrix}=-(a^2+b^2+c^2+d^2)^2.$$
Questions: For $n$ even, is the maximal dimension always $n$? Do similar such matrices exist with determinant $\pm\,(a_1^2+\cdots+a_n^2)^{n/2}$?
This is an attempt to generalize problem $3$ of the PUMA 2016.