We can easily extend our define of linear combination and span to general vector spaces:[br][br][u]Definition[/u]: For vectors [math]v_1,v_2, \ldots, v_n[/math] in [math]V[/math] and real numbers [math]c_1,c_2,\ldots, c_n[/math], [math]c_1v_1+c_2v_2+\cdots+c_nv_n[/math] is called a [b]linear combination[/b] of [math]v_1,v_2, \ldots, v_n[/math] with [b]weights[/b] numbers [math]c_1,c_2,\ldots, c_n[/math][br][br][u]Definition[/u]: [math]\text{Span}\left\{v_1,v_2, \ldots, v_n\right\}=\left\{c_1v_1+c_2v_2+\cdots+c_nv_n\in V \ | \ c_1,c_2,\ldots, c_n\in\mathbb{R}\right\}[/math] i.e. the set of all linear combinations of [math] v_1,v_2, \ldots, v_n[/math].[br][br][u]Remark[/u]: More generally, for any set [math]S[/math] of vectors in V (possibly infinite set), [math]\text{Span} \ S[/math] is the set of all linear combinations of any finite number of vectors in [math]S[/math].[br][br][u]Theorem[/u]: [math]\text{Span}\left\{v_1,v_2, \ldots, v_n\right\}[/math] is a subspace of [math]V[/math].[br][br]Proof: We just need to verify the three conditions:[br][list][*]The zero vector is a linear combination of [math] v_1,v_2, \ldots, v_n[/math].[/*][*]If [math]u[/math] and [math]v[/math] are linear combinations of [math] v_1,v_2, \ldots, v_n[/math], then so is [math]u+v[/math].[/*][*]If [math]c[/math] is any real number and [math]u[/math] is a linear combination of [math] v_1,v_2, \ldots, v_n[/math], so is [math]cu[/math]. [/*][/list][br][br]Also, the definition of linear independence is essentially the same as before:[br][br][u]Definition[/u]: The set [math]\left\{v_1,v_2, \ldots, v_n\right\}[/math] in [math]V[/math] is said to be [b]linearly independent[/b] if [math]c_1v_1+c_2v_2+\cdots+c_nv_n=0[/math] implies [math]c_1=c_2=\cdots=c_n=0[/math]. The set is said to be [b]linearly dependent[/b] if it is not linearly independent.[br][br][u]Remark[/u]: More generally, for any set [math]S[/math] of vectors in [math]V[/math] (possibly infinite set), [math]S[/math] is said to be linearly independent if any finite subset of [math]S[/math] is linearly independent.[br][br]The following is an equivalent definition of linearly dependence: [br][br][u]Theorem[/u]: A set of vectors [math]\left\{v_1,v_2,\ldots,v_n\right\}[/math] in a vector space [math]V[/math] is linearly dependent if and only if some [math]v_j[/math] is a linear combination of the others.[br][br]We already proved the theorem for [math]\mathbb{R}^n[/math]. The same proof also works for general vector spaces.[br][br][br][u]Example 1[/u]: Consider the set of polynomials [math]S=\{1,t,t^2\}[/math] in [math]\mathbb{P}[/math]. Show that [math]S[/math] is a linearly independent set and find [math]\text{Span} \ S[/math].[br][br]Linear independence: Suppose [math]c_1,c_2,c_3[/math] are real numbers such that [math]c_1\cdot 1+c_2t+c_3t^2=0[/math]. It is obvious that [math]c_1=c_2=c_3=0[/math]. Hence [math]S[/math] is a linear independent set.[br][br][math]\text{Span} \ S=\left\{c_1\cdot 1+c_2t+c_3t^2\in \mathbb{P} \ | \ c_1,c_2,c_3\in \mathbb{R}\right\}=\mathbb{P}_2[/math][br][br][br][u]Example 2[/u]: Consider the set [math]T=\left\{\begin{pmatrix}1&1\\0&0\end{pmatrix},\begin{pmatrix}0&1\\0&1\end{pmatrix},\begin{pmatrix}1&0\\0&-1\end{pmatrix}\right\}[/math] in [math]M_{2\times 2}[/math]. Is the set linearly independent?[br][br]To check whether the set is linearly independent, we need to consider the following equation:[br][br][math]c_1 \begin{pmatrix}1&1\\0&0\end{pmatrix}+c_2 \begin{pmatrix}0&1\\0&1\end{pmatrix}+c_3 \begin{pmatrix}1&0\\0&-1\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}[/math][br][math]\Rightarrow \begin{pmatrix}c_1+c_3&c_1+c_2\\0&c_2-c_3\end{pmatrix}= \begin{pmatrix}0&0\\0&0\end{pmatrix}[/math][br][br]And it is equivalent to solving the linear system [math]\left\{\begin{eqnarray}c_1+c_3 & = & 0 \\ c_1+c_2 & = & 0\\ c_2-c_3 & = & 0\end{eqnarray}\right.[/math]. It is not hard to see that the system is consistent and has nontrivial solutions like [math]c_1=1,c_2=-1,c_3=-1[/math]. That is to say, the weights of the linear combination are not all zero, which implies that the set is [u]linearly dependent[/u].[br][br][br]
Is the set [math]\left\{1+t,1+t+2t^2,3t-t^2\right\}[/math] in [math]\mathbb{P}[/math] a linearly independent set? Explain your answer briefly.
To check whether [math]\left\{1+t,1+t+2t^2,3t-t^2\right\}[/math] is linearly independent, we consider the following equation in [math]c_1,c_2[/math] and [math]c_3[/math]:[br][br][math]c_1(1+t)+c_2(1+t+2t^2)+c_3(3t-t^2)=0[/math][br][br]If the equation only has the trivial solution i.e. [math]c_1=c_2=c_3=0[/math], then the set is linearly independent. Simplifying the left-hand side of the equation, we get[br][br][math](2c_2-c_3)t^2+(c_1+c_2+3c_3)t+(c_1+c_2)=0[/math]. By comparing coefficients on both sides, we have the following system of linear equations in [math]c_1,c_2[/math] and [math]c_3[/math]:[br][br][math]\left\{\begin{eqnarray}2c_2-c_3 & = & 0\\ c_1+c_2+3c_3 & = & 0\\ c_1+c_2 \quad \quad & = & 0\end{eqnarray}\right.[/math][br][br]It can be shown that this linear system has only the trivial solution. Therefore, the set is linearly independent.
In [math]\mathbb{S}[/math], let [math]v_1[/math] be the sequence [math]1,0,0,0,\ldots [/math], [math]v_2[/math] be the sequence [math]0,1,0,0,\ldots[/math] and so on. What is [math]\text{Span} \left\{v_1,v_2,v_3, \ldots\right\}[/math]?
[u]Claim[/u]: [math]\text{Span} \left\{v_1,v_2,v_3, \ldots\right\}=\left\{(a_n)\in\mathbb{S} \ | \ a_n=0 \ \text{for all but finitely many} \ n \right\}[/math][br][br]For any [math](a_n)\in\text{Span} \left\{v_1,v_2,v_3, \ldots\right\}[/math], there exists a positive integer [math]N[/math] such that [math](a_n)=c_1v_1+c_2v_2+\cdots+c_Nv_N[/math]. Then [math]a_n=0[/math] for [math]n>N[/math] i.e. [math](a_n)\in \left\{(a_n)\in\mathbb{S} \ | \ a_n=0 \ \text{for all but finitely many} \ n \right\}[/math].[br][br]For any [math](a_n)\in \left\{(a_n)\in\mathbb{S} \ | \ a_n=0 \ \text{for all but finitely many} \ n \right\}[/math], there exists a positive integer [math]N[/math] such that [math]a_n=0[/math] for [math]n>N[/math]. Hence [math](a_n)=a_1v_1+a_2v_2+\cdots+a_Nv_N[/math], which implies that [br][math](a_n)\in\text{Span} \left\{v_1,v_2,v_3, \ldots\right\}[/math].