Is the vector [3 1] in Span {[1 2],[-1 1]}?

Question

Is the vector [latex]\left [ \begin{matrix} 3 \ 1 \end{matrix} ight ] [/latex] in Span [latex]\left\{\left [ \begin{matrix} 1 \ 2 \end{matrix} ight ] ,\left [ \begin{matrix} -1 \ 1 \end{matrix} ight ] ight\} [/latex] ?

Accepted Answer

Using the definition of span, the vector [latex]\left [ \begin{matrix} 3 \ 1 \end{matrix} ight ] [/latex] is in the spanned set if it can be written as a linear combination of the vectors in the spanning set. That is, we need to determine whether there exists [latex]c_{1},c_{2} \in \mathbb{R} [/latex] such that

[latex]\left [ \begin{matrix} 3 \ 1 \end{matrix} ight ] =c_{1}\left [ \begin{matrix} 1 \ 2 \end{matrix} ight ] +c_{2}\left [ \begin{matrix} -1 \ 1 \end{matrix} ight ] [/latex]

Performing operations on vectors on the right-hand side gives

[latex]\left [ \begin{matrix} 3 \ 1 \end{matrix} ight ] =\left [ \begin{matrix} c_{1} -c_{2} \ 2c_{1}+c_{2} \end{matrix} ight ] [/latex]

Since vectors are equal if and only if their corresponding entries are equal, we get that this vector equation implies

[latex]3=c_{1}-c_{2} \ 1=2c_{1}+c_{2}[/latex]

Adding the equations gives [latex]4=3c_{1}[/latex] and so [latex]c_{1}=\frac{4}{3}[/latex] .

Substituting this into either equation gives [latex]c_{2}=-\frac{5}{3} [/latex]. Hence, we have that

[latex]\left [ \begin{matrix} 3 \ 1 \end{matrix} ight ] =\frac{4}{3}\left [ \begin{matrix} 1 \ 2 \end{matrix} ight ] -\frac{5}{3}\left [ \begin{matrix} -1 \ 1 \end{matrix} ight ] [/latex]

Thus, by definition, [latex]\left [ \begin{matrix} 3 \ 1 \end{matrix} ight ] \in span \left\{\left [ \begin{matrix} 1 \ 2 \end{matrix} ight ] ,\left [ \begin{matrix} -1 \ 1 \end{matrix} ight ] ight\} [/latex].

Question 1.2.2: Is the vector [3 1] in Span {[1 2],[-1 1]}?

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