5.2 Vector Subspaces
Definition 5.2.1. A non-empty subset \(U\) of a vector space \(V\) is called a subspace of \(V\) if \(U\) is itself a vector space with the same operations of additions and scalar multiplication as those of \(V\).
Definition 5.2.2. Let \(V\) be a vector space over a field \(\mathbb {F}\). A non-empty subset \(U\) of \(V\) is a subspace of \(V\) if and only if \(\alpha u+v\in U\) for all \(\alpha \in \mathbb {F}\) and \(u,v\in U\).
Proof. Suppose that \(U\) is a subspace of \(V\). Then \(\alpha u\in U\) and so \(\alpha u+v\in U\).
Conversely, suppose that \(u,v,w\in U\) and \(\alpha , \beta \in \mathbb {F}\). Since the operations of \(U\) are those of \(V\) restricted to \(U\), \((u+v)+w=u+(v+w)\), \(u+v=v+u\), \(\alpha (u+v)=\alpha u+\alpha v\), \((\alpha +\beta )u=\alpha u+\beta u\) and \(1\cdot u=u\). It remains to show that \(U\) is closed under addition and scalar multiplication, \(0\in U\) guaranteed by the fact that \(\alpha u+v\in U\) for all \(u,v\in U\) and \(\alpha \in \mathbb {F}\). □
- 1.
- The subsets \(\{0\}\) and \(V\) are always subspaces of any vector space \(V\). They are called trivial subspaces. Any other subspace of \(V\) is a proper subspace.
- 2.
- The set \(U=\Big \{(x,0,0)\in \mathbb {R}^3:x\in \mathbb {R}\Big \}\) is a subspace of \(\mathbb {R}^3\).
- 3.
- The set \(U=\{x+iy:x,y\in \mathbb {Z}\}\) is not a subspace of \(\mathbb {C}\).
- 4.
- The set \(U=\Bigg \{ \begin {pmatrix} x&y\\-y&x\\ \end {pmatrix} \in M_{22}(\mathbb {R}):x,y\in \mathbb {R}\Bigg \}\) is a subspace of \(M_{22}(\mathbb {R})\).
Theorem 5.2.4. Let \(V\) be a vector space over a field \(\mathbb {F}\) and let \(U\) and \(W\) be subspaces of \(V\). Then \(U\cap W\) and \(U+W=\{u+w:u\in U\) and \(w\in W\}\) are both subspaces of \(V\).
Proof. Since both \(U\) and \(W\) are subspaces of \(V\) we have that \(0\in U\) and \(0\in W\), so that \(0\in U\cap W\). Thus \(U\cap W\neq \emptyset \). Now let \(\alpha \in \mathbb {F}\) and \(u,v\in U\cap W\). Since \(U\) and \(W\) are subspaces it follows that \(\alpha u+v\in U\) and \(\alpha u+v\in W\) so that \(\alpha u+v\in U\cap W\).
To prove the second part, let \(\alpha \in \mathbb {F}\) and \(u,v\in U+W\). Then there exists elements \(u_1,u_2\in U\) and \(w_1,w_2\in W\) such that \(u=u_1+w_1\) and \(v=u_2+w_2\). It follows
that
\(\alpha u+v=\alpha (u_1+w_1)+u_2+w_2=(\alpha u_1+u_2)+(\alpha w_1+w_2)\in U+W\) since \(U\) and \(W\) are subspaces and so \(\alpha u_1+u_2\in U\) and \(\alpha w_1+w_2\in W\).
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