2  Vector spaces, span, independence and basis

2.1 Vector spaces and subspaces

A vector space \(V\) over a field \(\mathbb F\) has vector addition and scalar multiplication satisfying the usual algebraic laws: addition is associative and commutative, has a zero vector and additive inverses, scalar multiplication respects multiplication and the identity scalar, and both distributive laws hold. Examples include \(\mathbb R^n\), matrices of a fixed size, polynomials, and real-valued functions on a fixed set with pointwise operations.

A subset \(W\subseteq V\) is a subspace if it is itself a vector space under the inherited operations. The subspace test says that \(W\) is a subspace exactly when it contains \(0\) and is closed under addition and scalar multiplication. The other axioms are inherited from \(V\).

For a linear system, \(\{x:Ax=0\}\) is a subspace: \(A(\alpha u+\beta v)=\alpha Au+\beta Av=0\) whenever \(Au=Av=0\). By contrast, \(\{(x,y):x+y=1\}\) is not a subspace, since it excludes \(0\).

2.2 Span and independence

A linear combination of \(v_1,\ldots,v_k\) is \(a_1v_1+\cdots+a_kv_k\) with scalars \(a_i\in\mathbb F\). Their span is the set of all such combinations. It is the smallest subspace containing these vectors: closure follows from combining coefficients, and any containing subspace must contain every combination.

The list \(v_1,\ldots,v_k\) is linearly independent if \[a_1v_1+\cdots+a_kv_k=0\quad\Longrightarrow\quad a_1=\cdots=a_k=0.\] Otherwise it is dependent. A zero vector makes any containing list dependent. For columns of a matrix, independence means the associated homogeneous system has only the zero solution, so elimination tests independence.

2.2.1 Worked example: a basis for a plane

Let \(W=\{(x,y,z)\in\mathbb R^3:x+y+z=0\}\). Set \(y=s\) and \(z=t\). Then \[\begin{pmatrix}x\\y\\z\end{pmatrix} =s\underbrace{\begin{pmatrix}-1\\1\\0\end{pmatrix}}_{u} +t\underbrace{\begin{pmatrix}-1\\0\\1\end{pmatrix}}_{v}.\] Thus \(u,v\) span \(W\). If \(au+bv=0\), its second coordinate gives \(a=0\) and its third gives \(b=0\), so they are independent. They form a basis. The vector \((-5,2,3)\) has coordinates \((2,3)\) in the ordered basis \((u,v)\).

2.3 Basis and dimension

A basis is a linearly independent spanning list (or, more generally, set). For an ordered finite basis \(B=(b_1,\ldots,b_n)\), every \(v\in V\) has unique coefficients: \[v=c_1b_1+\cdots+c_nb_n,\qquad [v]_B=(c_1,\ldots,c_n)^T.\] Spanning gives existence. If two expressions represented \(v\), subtracting them would produce a dependence relation, so independence gives uniqueness.

Basis size theorem. In a finite-dimensional vector space, every basis has the same number of vectors, called \(\dim V\). More precisely, any independent list has at most as many vectors as any finite spanning list.

Exchange argument. Express the first independent vector using a spanning list and replace a spanning vector with a nonzero coefficient by it. The new list still spans. Repeat with successive independent vectors. Independence ensures each step can replace a vector not already inserted. One cannot insert more independent vectors than the original list’s length. Apply this comparison in both directions to two bases.

This argument also yields useful constructions: remove redundant vectors from a finite spanning list to obtain a basis; extend an independent list to a basis in a finite-dimensional space. The zero space has the empty basis and dimension zero.

2.3.1 Worked example: polynomials are vectors too

Let \(\mathcal P_2=\{a+bx+cx^2:a,b,c\in\mathbb R\}\). The ordered list \(B=(1,x,x^2)\) is a basis, because the coefficients exist by definition and are unique. Hence \(\dim\mathcal P_2=3\) and \([2-x+3x^2]_B=(2,-1,3)^T\).

Consider \(C=(1,1+x,x+x^2)\). If \[a+b(1+x)+c(x+x^2)=0,\] comparison of \(x^2\), \(x\) and constant coefficients successively gives \(c=b=a=0\). Three independent vectors in the three-dimensional space form a basis. Solving \[2-x+3x^2=a+b(1+x)+c(x+x^2)\] gives \(c=3\), \(b=-4\), \(a=6\), so its \(C\)-coordinates are \((6,-4,3)^T\).

2.4 Column spaces and choosing original vectors

The column space of \(A\) is the span of its columns. To choose a basis from these columns, find pivot columns by row reduction, then select the corresponding original columns of \(A\). Row operations preserve dependencies among columns because they multiply on the left by an invertible matrix. They generally change the column space itself, so using the reduced columns as a basis for the original column space can be wrong.

2.5 Exercises

  1. Is \(W=\{(x,y):xy=0\}\) a subspace of \(\mathbb R^2\)?
  2. Find the dimension and a basis of \(\{(x,y,z):x=2y,\ z=0\}\).
  3. Are \((1,2)\), \((2,4)\) independent? What is their span’s dimension?
  4. Find the coordinates of \(3+2x\) in the basis \((1+x,1-x)\) of \(\mathcal P_1\).
  1. No. It contains \((1,0)\) and \((0,1)\) but not their sum \((1,1)\).
  2. Every vector is \(t(2,1,0)\), so \(\{(2,1,0)\}\) is a basis and the dimension is \(1\).
  3. They are dependent because the second is twice the first. Their span has dimension \(1\).
  4. Solve \(a+b=3\), \(a-b=2\): the coordinates are \((5/2,1/2)^T\).

Next: linear maps and coordinate matrices.