All Subspaces of $\mathbb{R^4}$ and $\mathbb{C^n}$ [Strang P129 3.1.14] 
The subspaces of $\mathbb{R^n}$  are $\mathbb{R^4}$  itself, three-dimensional planes $\mathbf{n \cdot v = 0}$, two-dimensional
  subspaces $\mathbf{n_1 \cdot v = 0}$ and $\mathbf{n_2 \cdot v = 0}$, one-dimensional lines through $(0, 0, 0, 0)$, and
  $(0, 0, 0, 0)$ by itself.



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*Does finding all subspaces of $\mathbb{R^n}$ involve the same level of difficulty? What about for $\mathbb{C^n}$?
If so, please overlook $\mathbb{R^4}$ herein and regard this question for $\mathbb{C^n}$.

*I recollect that $(0, 0, 0, 0)$ and $\mathbb{R^4}$ are the trivial subspaces of $\mathbb{R^4}$, but I don't perceive how to derive/deduce the other $3$ subspaces? 

*Would someone please explain the intuition behind the proper (nontrivial) subspaces? 
 A: 
Does finding all subspaces of $\Bbb R^n$ involve the same level of difficulty?
I don't perceive how to derive/deduce the other 3 subspaces.

Keep in mind that for any field $\Bbb F$, $k$ linearly independent vectors in $\Bbb F^n$ (with $k\leq n$) will span a $k$ dimensional subspace. So, it is possible to produce a $k$ dimensional subspace for every $k$ where $0\leq k\leq n$. This is true for any field: $\Bbb R$ or $\Bbb C$ are just particular cases.
Secondly, "the other 3 subspaces" is not really an accurate thing to say... there are many subspaces of each dimension. For example, you can put a set of $1$-dimensional subspaces into correspondence with the nonzero elements of $\Bbb F$, so if $F$ is infinite, there are infinitely many distinct $1$-dimensional subspaces.
But yes, the subspaces necessarily have dimension between $0$ and $n$.

Would someone please explain the intuition behind the proper (nontrivial) subspaces?

They are just vector spaces of smaller dimension contained in $\Bbb F^n$, and they use the same addition and scalar multiplication as the containing vector space. What else is there to say?
