Vtyjkyuk

I am studying linear algebra.
I found this exercise
Let V be a subspace of all n X n square matrices such that every nonzero element of V is invertible. Show that V is atmost n dimensional. Also prove that if n is odd then dimV =1.

I am unable to approach that exactly which type of matrices will be there in V?

Assume that $K$ is the base field. Let $m:=dim_K(V)$ and denote by $leftA_1,A_2,ldots,A_mright$ a basis of $V$. Note that the first columns of the matrices $A_i$'s must be linearly independent; otherwise, there are scalars $a_1,a_2,ldots,a_m$, not all being zero, such that the first column of $sumlimits_i=1^m,a_iA_i$ is zero. Since all nonzero elements of $V$ are invertible, we must have
$$sumlimits_i=1^m,a_iA_i=0,,$$ which contradicts the assumption that the $A_i$'s form a basis of $V$. As the first columns of the matrices $A_i$'s are linearly independent elements of $K^n$, we conclude that $mleq n$.

If $n$ is odd and $K:=mathbbR$, then note that, for any two invertible matrices $A$ and $B$, there exists $lambdainmathbbR$ such that $A-lambda,B$ is singular. This is because $A^-1B$ has a real eigenvalue and we can take $lambda$ to be such an eigenvalue. This proves that $mleq 1$ when $n$ is odd and $K$ is the field of real numbers. The same argument also shows that, if $K$ is algebraically closed (and $n$ is an arbitrary positive integer), then $mleq 1$.