Question

In Exercises 5-8, find the coordinate vector \({\left[ {\mathop{\rm x}\nolimits} \right]_B}\) of \({\mathop{\rm x}\nolimits} \) relative to the given basis \(B = \left\{ {{{\mathop{\rm b}\nolimits} _1},...,{{\mathop{\rm b}\nolimits} _n}} \right\}\).

8. \({{\mathop{\rm b}\nolimits} _1} = \left[ {\begin{array}{*{20}{c}}1\\0\\3\end{array}} \right],{{\mathop{\rm b}\nolimits} _2} = \left[ {\begin{array}{*{20}{c}}2\\1\\8\end{array}} \right],{{\mathop{\rm b}\nolimits} _3} = \left[ {\begin{array}{*{20}{c}}1\\{ - 1}\\2\end{array}} \right],{\mathop{\rm x}\nolimits} = \left[ {\begin{array}{*{20}{c}}3\\{ - 5}\\4\end{array}} \right]\)

Short answer

The coordinate vector \({\left[ {\mathop{\rm x}\nolimits} \right]_B}\) of x relative to the given basis is \({\left[ {\mathop{\rm x}\nolimits} \right]_B} = \left[ {\begin{array}{*{20}{c}}{ - 2}\\0\\5\end{array}} \right]\).

Step-by-step solution

Define the coordinate vector of x

Suppose \(B = \left\{ {{{\mathop{\rm b}\nolimits} _1},...,{{\mathop{\rm b}\nolimits} _n}} \right\}\) is a basis for \(V\) and x is in \(V\). Thecoordinatesof \({\mathop{\rm x}\nolimits} \) relative to basis \(B\) (or the \(B\)-coordinates of x) are the weights \({c_1},...,{c_n}\), such that \({\mathop{\rm x}\nolimits} = {c_1}{b_1} + ... + {c_n}{b_n}\).

Write an augmented matrix

Consider the augmented matrix shown below.

\(\left[ {\begin{array}{*{20}{c}}{{{\mathop{\rm b}\nolimits} _1}}&{{{\mathop{\rm b}\nolimits} _2}}&{{{\mathop{\rm b}\nolimits} _3}}&{\mathop{\rm x}\nolimits} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&2&1&3\\0&1&{ - 1}&{ - 5}\\3&8&2&4\end{array}} \right]\)

Apply the row operation

At row 3, multiply row 1 by 3 and subtract it from row 3. At row 1, multiply row 2 by 2 and subtract it from row 1.

\[ \sim \left[ {\begin{array}{*{20}{c}}1&0&3&{13}\\0&1&{ - 1}&{ - 5}\\0&2&{ - 1}&{ - 5}\end{array}} \right]\]

At row 3, multiply row 2 by 2 and subtract it from row 3.

\[ \sim \left[ {\begin{array}{*{20}{c}}1&0&3&{13}\\0&1&{ - 1}&{ - 5}\\0&0&1&5\end{array}} \right]\]

At row 1, multiply row 3 by 3 and subtract it from row 1. At row 2, multiply row 2 by 1 and add it to row 3.

\[ \sim \left[ {\begin{array}{*{20}{c}}1&0&0&{ - 2}\\0&1&0&0\\0&0&1&5\end{array}} \right]\]

Determine the coordinate vector \({\left[ {\mathop{\rm x}\nolimits}  \right]_B}\) of \({\mathop{\rm x}\nolimits} \)

The solution of matrix A is \({c_1} = - 2,{c_2} = 0,{c_3} = 5\).

The coordinates vector \({\left[ {\mathop{\rm x}\nolimits} \right]_B}\) of \({\mathop{\rm x}\nolimits} \) relative to the given basis is shown below:

\(\begin{array}{c}{\left[ {\mathop{\rm x}\nolimits} \right]_B} = \left[ {\begin{array}{*{20}{c}}{{c_1}}\\{{c_2}}\\{{c_3}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - 2}\\0\\5\end{array}} \right]\end{array}\)

Thus, the coordinates vector \({\left[ {\mathop{\rm x}\nolimits} \right]_B}\) of \({\mathop{\rm x}\nolimits} \) relative to the given basis is \({\left[ {\mathop{\rm x}\nolimits} \right]_B} = \left[ {\begin{array}{*{20}{c}}{ - 2}\\0\\5\end{array}} \right]\).