Suppose $V$ is a vector space such that the only subspaces of $V$ are $\{0\}$ and $V$. Determine the dimension of $V$. Suppose $V$ is a vector space such that the only subspaces of $V$ are $\{0\}$ and $V$. Determine the dimension of $V$.
I have no idea how to prove that. I know that $\{0\}$ and $V$ are always subspaces of $V$. That is we have to find subspace having no proper subspaces and find its dimension. How to get this. Please help.
 A: Let $v\in V$ be a non-zero vector. Then $\Bbb R\cdot v = \{r\cdot v: r\in\Bbb R\}$ is a subspace. Since it is clearly not the zero subspace, it is all of $V$ by assumption. So by definition $V$ has dimension $1$.
A: Consider $x\in V$. The subspace spanned by $x$ is
$$\langle x\rangle=\{y\in V,\, \exists \lambda\in \Bbb{K}\,y=\lambda x\}$$
Now by assumption $\langle x\rangle=\{0\}\text{ or }V$. So either $x=0$ or it spans the whole space and this means the dimension is $0$ (trivial case) or $1$.
A: The dimension is equal to 1. 
Recall that the dimension of a vector space $V$ is given by the number of basis vectors it has. Since $V$ is not the null-space we know the dimension is at least 1. Every finite dimensional vector space $V$ is isomorphic to $\mathbb{R}^n$. We also can always find a subspace of dimension $n-1$ by removing one basis vector. Since the only subspace (the zero space) has dimension 0, $V$ must have dimension $1$, because if it didn't, then we could find a $n-1$ dimensional subspace. 
