Problem related to find dimension of a subspace 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?  
 A: Assume that $K$ is the base field.  Let $m:=\dim_K(V)$ and denote by $\left\{A_1,A_2,\ldots,A_m\right\}$ 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 $\sum\limits_{i=1}^m\,a_iA_i$ is zero.  Since all nonzero elements of $V$ are invertible, we must have
$$\sum\limits_{i=1}^m\,a_iA_i=0\,,$$ 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 $m\leq n$.
If $n$ is odd and $K:=\mathbb{R}$, then note that, for any two invertible matrices $A$ and $B$, there exists $\lambda\in\mathbb{R}$ such that $A-\lambda\,B$ is singular.  This is because $A^{-1}B$ has a real eigenvalue and we can take $\lambda$ to be such an eigenvalue.  This proves that $m\leq 1$ 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 $m\leq 1$.
