Problems on vector spaces Let $E$ a $\mathbb{K}$-vector space of finite dimension $n$, $\mathcal{V}$ a subspace of $\mathcal{L}(E)$ such that $$\forall u\in\mathcal{V}\setminus \{0\},u\in\mathcal{GL}(E)$$
a) Show that $\dim\mathcal{V}\leq n$
b) When $\mathbb{K}=\mathbb{C}$, show that $\dim\mathcal{V}\leq1$
c) When $E=\mathbb{R}^2$, give an example of a space $\mathcal{V}$ with dimension $2$.
d) Suppose that $\mathbb{K}=\mathbb{R}$. Show that if $\dim\mathcal{V}\geq2$ then $\dim E$ is even.
EDIT : questions  a,b,c have been answered in the comments/chat. I still need help with d).

For c), I thought that $\mathcal{V}=\{u(x,y)=(a\cdot x+b\cdot y,a\cdot y-b\cdot x)\mid a,b\in\mathbb{R}\}$ worked well.
I however have not found, after searching some time, how to do a),b) and d) (which it would seem are independent).
Could I get some hints for those three questions?
 A: I present all answers; partly gleaned from the comments.
Part a:
Let $f:E \to \Bbb K$ be a non-zero linear functional.
Let $W$ be the set $v \in \mathcal L(E)$ such that $f \circ v = 0$.
Note that $W$ is $n^2-n$ dimensional.  So, if $V$ is any subspace of $\mathcal L(E)$ with dimension greater than $n$, then there is a $v \in V$ with $f \circ v = 0$.  
The conclusion follows.
Part b:
Consider $v,w \in \mathcal {GL}(E)$, and suppose that they are linearly independent.  We note that $\det(v - tw)$ is a polynomial on $t$.  By the fundamental theorem of algebra, there is a $t \in \Bbb C$ with $\det(v - tw) = 0$.  Thus, $v - tw$ is not invertible.  However, since $v,w$ are linearly independent, $v - tw \neq 0$.
It follows that no two-dimensional subspace of $\mathcal{L}(E)$ can satisfy the properties of $V$.
Part c:
Consider
$$
\left\{a \pmatrix{1&0\\0&1} + b\pmatrix{0&-1\\1&0}:a,b \in \Bbb R \right\}
$$
Part d:
Suppose that $E$ has odd degree.  Consider $v,w \in \mathcal{GL}(E)$.  We note that $\det(v - tw)$ is a polynomial on $t$ of odd degree.  Noting that any polynomial of odd degree has a real root, apply the argument from b.
