Prove that the only subspaces of \(\RR^1\) are \(\RR^1\) and the zero subspace.
Prove that a subspace of \(\RR^2\) is \(\RR^2\), or the zero subspace, or consists of all scalar multiples of some fixed vector in \(\RR^2\). (The last type of subspace is, intuitively, a straight line through the origin.)
Can you describe the subspaces of \(\RR^3\)?
Solution
1. The only subspaces of \(\RR\) are \(\{0\}\) and \(\RR\).
Let \(W\le\RR\). If \(W=\{0\}\) there is nothing to prove. Otherwise choose \(0\ne a\in W\). For every \(x\in\RR\), \[ x=\frac{x}{a}a\in W, \] so \(W=\RR\).
2. Every subspace of \(\RR^2\) is either \(\{0\}\), a line through the origin, or all of \(\RR^2\).
Let \(W\le\RR^2\). If \(W=\{0\}\), done. Otherwise choose \(0\ne v\in W\). If every vector of \(W\) is a scalar multiple of \(v\), then \(W=\operatorname{span}\{v\}\), a line through the origin. Otherwise choose \(w\in W\) not proportional to \(v\). Then \(v,w\) are linearly independent and hence form a basis of \(\RR^2\), so \(W=\RR^2\).
3. Every subspace of \(\RR^3\) is one of \(\{0\}\), a line through the origin, a plane through the origin, or all of \(\RR^3\).
A subspace has dimension \(0,1,2,\) or \(3\). Dimension \(0\) gives \(\{0\}\); dimension \(1\) gives the span of one nonzero vector, hence a line through the origin; dimension \(2\) gives the span of two independent vectors, hence a plane through the origin; dimension \(3\) gives \(\RR^3\).