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Mathematics LibreTexts

5.1: Linear Span

  • Page ID
    289
  • [ "article:topic", "authorname:schilling", "span" ]

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    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: Linear Span

    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: Subspaces

    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 \(U\).

    \(\square\)

    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: finite-dimensional and Infinite-dimensional vector spaces

    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 \(\PageIndex{1}\):

    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 \(\PageIndex{2}\):

    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 \(\PageIndex{3}\):

    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 \(z^{m+1}\notin \Span(p_1(z),\ldots,p_k(z))\).