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Mathematics LibreTexts

5.1 Linear span

As before, let \(V\) denote a vector space over \(\mathbb{F}\). Given vectors \(v_1,v_2,\ldots,v_m\in V\), a vector \(v\in V\) is a linear combination of \((v_1,\ldots,v_m)\) if there exist scalars \(a_1,\ldots,a_m\in\mathbb{F}\) such that

\[ v = a_1 v_1 + a_2 v_2 + \cdots + a_m v_m.\]

Definition 5.1.1.  The linear span (or simply span) of \((v_1,\ldots,v_m)\) is defined as

\[ \Span(v_1,\ldots,v_m) := \{ a_1 v_1 + \cdots + a_m v_m \mid a_1,\ldots,a_m \in \mathbb{F} \}.\]

Lemma 5.1.2. Let \(V\) be a vector space and \(v_1,v_2,\ldots,v_m\in V\). Then

  1. \(v_j\in \Span(v_1,v_2,\ldots,v_m)\).
  2. \(\Span(v_1,v_2,\ldots,v_m)\) is a subspace of \(V\).
  3. If \(U\subset V\) is a subspace such that \(v_1,v_2,\ldots v_m\in U\), then \(\Span(v_1,v_2,\ldots,v_m)\subset U\).

Proof.  Property~1 is obvious. For Property~2, note that \(0\in\Span(v_1,v_2,\ldots,v_m)\) and that \(\Span(v_1,v_2,\ldots,v_m)\) is closed under addition and scalar multiplication.  For Property~3, note that a subspace \(U\) of a vector space \(V\) is closed under addition and scalar multiplication. Hence, if \(v_1,\ldots,v_m\in U\), then any linear combination \(a_1v_1+\cdots +a_m v_m\) must also be an element of \(U\).

         Lemma 5.1.2 implies that \(\Span(v_1,v_2,\ldots,v_m)\) is the smallest subspace of \(V\) containing each of \(v_1,v_2,\ldots,v_m\).

Definition 5.1.3. If \(\Span(v_1,\ldots,v_m)=V\), then we say that \((v_1,\ldots,v_m)\) spans \(V\) and we call \(V\) finite-dimensional. A vector space that is not finite-dimensional is called infinite-dimensional.

Example 5.1.4.  The vectors \(e_1=(1,0,\ldots,0)\), \(e_2=(0,1,0,\ldots,0), \ldots, e_n=(0,\ldots,0,1)\) span \(\mathbb{F}^n\). Hence \(\mathbb{F}^n\) is finite-dimensional.

Example 5.1.5.  The vectors \(v_1=(1,1,0)\) and \(v_2=(1,-1,0)\) span a subspace of \(\mathbb{R}^3\). More precisely, if we write the vectors in \(\mathbb{R}^3\) as 3-tuples of the form \((x,y,z)\), then \(\Span(v_1,v_2)\) is the \(xy\)-plane in \(\mathbb{R}^3\). 

Example 5.1.6. Recall that if \(p(z)=a_mz^m + a_{m-1} z^{m-1} + \cdots + a_1z + a_0\in \mathbb{F}[z]\) is a polynomial with coefficients in \(\mathbb{F}\) such that \(a_m\neq 0\), then we say that \(p(z)\) has degree \(m\). By convention, the degree of the zero polynomial \(p(z)=0\) is \(-\infty\). We denote the degree of \(p(z)\) by \(\deg(p(z))\). Define

\( \mathbb{F}_m[z] = \) set  of all polynomials in \( \mathbb{F}[z] \) of degree at most m.

Then \(\mathbb{F}_m[z]\subset \mathbb{F}[z]\) is a subspace since \(\mathbb{F}_m[z]\) contains the zero polynomial and is closed under addition and scalar multiplication. In fact, \(\mathbb{F}_m[z]\) is a finite-dimensional subspace of \(\mathbb{F}[z]\) since

\[ \mathbb{F}_m[z] = \Span(1,z,z^2,\ldots,z^m). \]

At the same time, though, note that \(\mathbb{F}[z]\) itself is infinite-dimensional. To see this, assume the contrary, namely that

\[ \mathbb{F}[z] = \Span(p_1(z),\ldots,p_k(z))\]

for a finite set of \(k\) polynomials \(p_1(z),\ldots,p_k(z)\). Let \(m=\max(\deg p_1(z),\ldots,\deg p_k(z))\). Then \(z^{m+1}\in\mathbb{F}[z]\), but \(z^{m+1}\notin \Span(p_1(z),\ldots,p_k(z))\).