The set \(B = \left\{ {1 - {t^2},t - {t^2},2 - 2t + {t^2}} \right\}\) is a basis for \({{\mathop{\rm P}\nolimits} _2}\). Find the coordinate vector of \({\mathop{\rm p}\nolimits} \left( t \right) = 3 + t - 6{t^2}\) relative to \(B\).

Answer:

The coordinate vector of \({\mathop{\rm p}\nolimits} \left( t \right) = 3 + t - 6{t^2}\) relative to \(B\) is \({\left[ {\mathop{\rm p}\nolimits} \right]_B} = \left[ {\begin{array}{*{20}{c}}7\\{ - 3}\\{ - 2}\end{array}} \right]\).

Determine \({c_1},{c_2}\), and \({c_3}\) as shown below:

\[{c_1}\left( {1 - {t^2}} \right) + {c_2}\left( {t - {t^2}} \right) + {c_3}\left( {2 - 2t + {t^2}} \right) = {\mathop{\rm p}\nolimits} \left( t \right) = 3 + t - 6{t^2}\]

Equate the coefficients of the two polynomials to obtain the system of equations as shown below:

\[\begin{array}{c}{c_1} + \,\,\,\,\,\,\,2{c_3} = 3\\\,\,\,\,\,\,\,\,{c_2} - 2{c_3} = 1\\ - {c_1} - {c_2} + {c_3} = - 6\end{array}\]

Convert the system of equations into an augmented matrix, as shown below:

\(\left[ {\begin{array}{*{20}{c}}1&0&2&3\\0&1&{ - 2}&1\\{ - 1}&{ - 1}&1&{ - 6}\end{array}} \right]\)

At row 3, add rows 1 and 3.

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

At row 3, add rows 2 and 3.

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

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

\( \sim \left[ {\begin{array}{*{20}{c}}1&0&0&7\\0&1&{ - 2}&1\\0&0&1&{ - 2}\end{array}} \right]\)

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

\( \sim \left[ {\begin{array}{*{20}{c}}1&0&0&7\\0&1&0&{ - 3}\\0&0&1&{ - 2}\end{array}} \right]\)

The coordinate vector of \({\mathop{\rm p}\nolimits} \left( t \right) = 3 + t - 6{t^2}\) is \({\left[ {\mathop{\rm p}\nolimits} \right]_B} = \left[ {\begin{array}{*{20}{c}}7\\{ - 3}\\{ - 2}\end{array}} \right]\).

It is also possible to solve this problem using the coordinate vectors of the given polynomials relative to the standard basis \(\left\{ {1,t,{t^2}} \right\}\). The same system of equations will be obtained in the result.

Thus, the coordinate vector of \({\mathop{\rm p}\nolimits} \left( t \right) = 3 + t - 6{t^2}\) relative to \(B\) is \({\left[ {\mathop{\rm p}\nolimits} \right]_B} = \left[ {\begin{array}{*{20}{c}}7\\{ - 3}\\{ - 2}\end{array}} \right]\).