The first question in linear algebra is not how to compute with matrices, but what kind of objects admit addition and scalar multiplication. Vector spaces isolate exactly the algebraic rules needed for linear combination arguments.
Vector space axioms and first examples
Definition. Let \(F\) be a field. A vector space over \(F\) is a set \(V\) with addition \((u,v)\mapsto u+v\) and scalar multiplication \((a,v)\mapsto av\), such that \((V,+)\) is an abelian group and, for all \(a,b\in F\) and \(u,v\in V\), \[ a(u+v)=au+av,\qquad (a+b)v=av+bv,\qquad a(bv)=(ab)v,\qquad 1v=v. \] The zero vector is denoted \(0\), and the additive inverse of \(v\) is denoted \(-v\).
Example. The standard examples are \(F^n\), the polynomial space \(F[x]\), the space \(F_m[x]\) of polynomials of degree at most \(m\), the matrix space \(M_{m,n}(F)\), the function space \(F^X=\{f:X\to F\}\), and solution spaces of homogeneous linear equations. The set of polynomials of degree exactly \(m\) is not a vector space, because adding two degree \(m\) polynomials can cancel the leading terms.
Subspaces and the subspace test
Definition. A subset \(W\subset V\) is a subspace if it is a vector space under the inherited operations. Equivalently, \(W\) is nonempty and satisfies \[ u,v\in W,\\ a,b\in F \quad\Longrightarrow\quad au+bv\in W. \] This one-line condition is the subspace test.
Proof. If \(W\) is closed under all linear combinations \(au+bv\), then taking \(a=b=1\) gives closure under addition, taking \(b=0\) gives closure under scalar multiplication, and taking \(a=0\) in a nonempty set gives the zero vector. The remaining vector space axioms are inherited from \(V\). Conversely, any subspace is closed under addition and scalar multiplication, hence under \(au+bv\).
Generated subspaces
Proposition. If \(S\subset V\), then \[ \operatorname{span}(S)=\left\{\sum_{i=1}^r a_i s_i:r\ge 0,\ a_i\in F,\ s_i\in S\right\} \] is the smallest subspace of \(V\) containing \(S\). Here the empty sum is \(0\).
Proof. The set of finite linear combinations is closed under linear combinations, so it is a subspace. Every subspace containing \(S\) must contain all finite linear combinations of elements of \(S\), so no smaller subspace containing \(S\) is possible.
Sums, intersections, and direct sums
Theorem. If \(U,W\subset V\) are subspaces, then \(U\cap W\) and \[ U+W=\{u+w:u\in U,\ w\in W\} \] are subspaces. The sum is direct, written \(U\oplus W\), exactly when every vector of \(U+W\) has a unique representation \(u+w\), equivalently \(U\cap W=\{0\}\).
Proof. Intersections are closed under linear combinations because membership in every subspace is preserved. Sums are closed because \(a(u_1+w_1)+b(u_2+w_2)=(au_1+bu_2)+(aw_1+bw_2)\). If \(u+w=u'+w'\), then \(u-u'=w'-w\in U\cap W\). Thus uniqueness is equivalent to the intersection being zero.
Affine subspaces and a common mistake
Counterexample. A line in \(\mathbb R^2\) not passing through the origin, such as \(\{(x,1):x\in\mathbb R\}\), is closed under neither scalar multiplication nor addition. It is an affine subspace, not a linear subspace. Linear subspaces always contain \(0\).
Remark. The habit to build is this: whenever a problem gives a set of vectors, first ask for the subspace it spans, the linear equations it satisfies, and how it sits inside sums and intersections with other subspaces. Those three views recur throughout the note.