5.3 Linear Combinations
Definition 5.3.1. Let \(V\) be a vector space over a field \(\mathbb {F}\) and \(S=\{v_1,v_2,\dots ,v_n\}\) a subset of \(V\). A vector \(v\in V\) is said to be a linear combination of the vectors in \(S\) if there exist scalars \(\alpha _1, \alpha _2,\dots ,\alpha _n\in \mathbb {F}\) such that \(v=\alpha _1v_1+\alpha _2v_2+\cdots +\alpha _nv_n\)
Theorem 5.3.2. Let \(V\) be a vector space over a field \(\mathbb {F}\) and \(U\) a subspace of \(V\). If \(u_1,u_2,\dots ,u_n\) are elements of \(U\), then \(\{\alpha _1 u_1+\alpha _2 u_2+\cdots +\alpha _nu_n:\alpha _i\in \mathbb {F}, i=1,2,\dots ,n\}\subseteq U\).
Proof. We prove by mathematical induction on \(n\). If \(n=1\), then \(\alpha _1u_1\in U\) by definition of a subspace. Suppose that every linear combination of \(n-1\) vectors in \(u\) is also an element of \(U\). Let \(u=\alpha _1u_1+\alpha _2u_2+\cdots +\alpha _nu_n\) be a linear combination of vectors in \(u_1\), clearly, the element \(v=\alpha _1u_2+\cdots +\alpha _nu_n\) is in \(u\) by the induction hypothesis. Since \(u=\alpha _1u_1+v\) and \(\alpha _1u_1,v\in U\), it follows from the fact that \(u\) is a subspace of \(V\) that \(\alpha _1u_1+v=u\in U.\) □
Definition 5.3.3. Let \(V\) be a vector space over a field \(\mathbb {F}\) and \(S=\{u_1,u_2,\dots ,u_n\}\) a subset of \(V\). The set Span\((S)=\{\alpha _1u_1+\alpha _2u_2+\cdots +\alpha _nu_n: \alpha _i\in \mathbb {F}\}\) is called the Span of \(S\).
If \(V=\) Span\((S)\), then we say that is Spanned or generated by \(S\).
- 1.
- The vector space \(\mathbb {R}^3\) is Spanned by the set \(S=\{(1,0,0), (0,1,0), (0,0,1)\}\) i.e.
\(\mathbb {R}^3=\) Span\(\{(1,0,0), (0,1,0), (0,0,1)\}\). - 2.
- The vector space \(\mathbb {R}^n\) is spanned by \(S=\{(1,0,\dots ,0), (0,1,\dots ,0),\dots , (0,0,\dots ,0,1)\}\).
Theorem 5.3.5. Let \(V\) be a vector space over a field \(\mathbb {F}\) and \(S=\{v_1,v_2,\dots ,v_n\}\). Then Span\((S)\) is a subspace of \(V\).
Proof. Let \(\alpha \in \mathbb {F}\) and \(v=\alpha _1v_1+\alpha _2v_2+\cdots +\alpha _nv_n\in \) Span\((S)\). Then \(\alpha v=\alpha (\alpha _1v_1+\alpha _2v_2+\cdots +\alpha _nv_n)=(\alpha \alpha _1)v_1+(\alpha \alpha _2)v_2+\cdots +(\alpha \alpha _n)v_n\in \) Span\((S)\). Since \(\alpha \alpha _1, \alpha \alpha _2,\dots ,\alpha \alpha _n\in \mathbb {F}\). Also if \(u=\beta _1v_1+\beta _2v_2+\cdots +\beta _nv_n\in \) Span\((S)\), then \((u+v)=(\alpha _1+\beta _1)v_1+(\alpha _2+\beta _2)v_2+\cdots +(\alpha _n+\beta _n)v_n\in \) Span\((S)\), since
\(\alpha _1+\beta _1, \alpha _2+\beta _2,\dots ,\alpha _n+\beta _n\in \mathbb {F}\). □
Definition 5.3.6. Let \(V\) be a vector space over a field \(\mathbb {F}\). A set \(S=\{v_1,v_2,\dots ,v_n\}\) of vectors in \(V\) is said to be linearly independent if whenever \(\alpha _1, \alpha _2,\dots ,\alpha _n\in \mathbb {F}\) are such that \(\alpha _1v_1+\alpha _2v_2+\cdots +\alpha _nv_n=0\), then \(\alpha _1=\alpha _2=\cdots =\alpha _n=0\).
A set which is not linearly independent is said to be linearly dependent.
Example 5.3.7. Determine whether the following set of vectors in \(\mathbb {R}\) is linearly independent;
\(S=\{(1,-1,-1,2), (2,-2,-1,3), (-1,1,-1,0)\}\).
Solution. Let \(v_1=(1,-1,-1,2), v_2=(2,-2,-1,3)\) and \(v_3=(-1,1,-1,0)\) and \(\alpha _1, \alpha _2,\alpha _3\in \mathbb {R}\) such that \(\alpha _1v_1+\alpha _2v_2+\alpha _3v_3=0=(0,0,0,0)\). Then we have
\((\alpha _1,-\alpha _1,-\alpha _1,2\alpha _1)+(2\alpha _2,-2\alpha _2,-\alpha _2,3\alpha _2)+(-\alpha _3,\alpha _3,-\alpha _3,0)=(0,0,0,0)\) which leads to the homogeneous system \begin {align*} \alpha _1+2\alpha _2-\alpha _3 &=0\\ -\alpha _1-2\alpha _2+\alpha _3 &=0\\ -\alpha _1-\alpha _2-\alpha _3 &=0\\ 2\alpha _1+3\alpha _2-0 &=0 \end {align*}
The coefficient matrix is \( \begin {pmatrix} 1&2&-1\\-1&-2&1\\-1&-1&-1\\2&3&0\\ \end {pmatrix} \)
Applying elementary row operations, we obtain
\begin {align*} \begin {pmatrix} 1&2&-1\\-1&-2&1\\-1&-1&-1\\2&3&0\\ \end {pmatrix} & \hspace {0.3cm} \begin {matrix} r_2\rightarrow r_2+r_1\\ \longrightarrow \\ r_3\rightarrow r_3+r_1\\ r_4\rightarrow r_4-2r_1\\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} 1&2&-1\\0&0&0\\0&1&-2\\0&-1&2\\ \end {pmatrix} \hspace {0.3cm} \begin {matrix} r_4\rightarrow r_4+r_3\\ \longrightarrow \\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} 1&2&-1\\0&0&0\\0&1&-2\\0&0&0\\ \end {pmatrix}\\\\ & \hspace {0.3cm} \begin {matrix} r_2\longleftrightarrow r_3\\ \longrightarrow \\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} 1&2&-1\\0&1&-2\\0&0&0\\0&0&0\\ \end {pmatrix} \end {align*}
This leads to \(\alpha _2=2\alpha _3\) so that \(\alpha _1+2\alpha _2-\alpha _3=0\) and \(\alpha _3=\alpha _1+2\alpha _2\). Thus this system has a non-trivial solution. For example, if \(\alpha _1=0\) and \(\alpha _2=1\), then \(\alpha _3=2\) and \((0,1,2)\) is not \((0,0,0)\).
Hence \(v_1, v_2, v_3\) are linearly dependent.
Example 5.3.8. Show that the set of vectors \(S=\{v_1=(2,1,2), v_2=(1,2,-2), v_3=(3,2,5)\}\) in \(\mathbb {R}^3\) are linearly independent.
Solution. Let \(\alpha _1, \alpha _2, \alpha _3\in \mathbb {R}\) be such that \(\alpha _1v_1+\alpha _2v_2+\alpha _3v_3=(0,0,0)\). Then we obtain the system \begin {align*} 2\alpha _1+\alpha _2+3\alpha _3 &=0\\ \alpha _1+2\alpha _2+2\alpha _3 &=0\\ 2\alpha _1-2\alpha _2+5\alpha _3 &=0 \end {align*}
The coefficient matrix is \( \begin {pmatrix} 2&1&3\\1&2&2\\2&-2&5\\ \end {pmatrix} \) and \begin {align*} \begin {pmatrix} 2&1&3\\1&2&2\\2&-2&5\\ \end {pmatrix} &\hspace {0.3cm} \begin {matrix} r_2\rightarrow r_2-r_1\\ \longrightarrow \\ r_3\rightarrow r_3-r_1\\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} 2&1&3\\-1&1&-1\\0&-3&-2\\ \end {pmatrix} \hspace {0.3cm} \begin {matrix} r_1\rightarrow r_1+2r_2\\ \longrightarrow \\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} 0&3&1\\-1&1&-1\\0&-3&2\\ \end {pmatrix}\\\\ &\hspace {0.3cm} \begin {matrix} r_3\rightarrow r_3+r_1\\ \longrightarrow \\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} 0&3&1\\-1&1&-1\\0&0&-1\\ \end {pmatrix} \hspace {0.3cm} \begin {matrix} r_1\longleftrightarrow r_2\\ \longrightarrow \\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} -1&1&-1\\0&3&1\\0&0&-1\\ \end {pmatrix}\\\\ & \hspace {0.3cm} \begin {matrix} r_1\rightarrow r_1-r_3\\ \longrightarrow \\ r_2\rightarrow r_2+r_3\\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} -1&1&0\\0&3&0\\0&0&-1\\ \end {pmatrix} \hspace {0.3cm} \begin {matrix} r_3\rightarrow \frac {1}{3}r_2\\ \longrightarrow \\ r_3\rightarrow -r_3\\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} -1&1&0\\0&1&0\\0&0&1\\ \end {pmatrix}\\\\ & \hspace {0.3cm} \begin {matrix} r_1\rightarrow r_1-r_2\\ \longrightarrow \\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} -1&0&0\\0&1&0\\0&0&1\\ \end {pmatrix} \hspace {0.3cm} \begin {matrix} r_1\rightarrow -r_1\\ \longrightarrow \\ \end {matrix} \hspace {0.3cm} \begin {pmatrix} 1&0&0\\0&1&0\\0&0&1\\ \end {pmatrix}. \end {align*}
Hence the trivial solution \((\alpha _1,\alpha _2, \alpha _3)=(0,0,0)\) is the unique solution. This means that the vectors \(v_1, v_2\) and \(v_3\) are linearly independent.
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