Question

The set \(B = \left\{ {1 + {t^2},t + {t^2},1 + 2t + {t^2}} \right\}\) is a basis for \({{\mathop{\rm P}\nolimits} _2}\). Find the coordinate vector of \({\mathop{\rm p}\nolimits} \left( t \right) = 1 + 4t + 7{t^2}\) relative to \(B\).

Answer:

The coordinate vector of \({\mathop{\rm p}\nolimits} \left( t \right) = 1 + 4t + 7{t^2}\) relative to \(B\) is \({\left[ {\mathop{\rm p}\nolimits} \right]_B} = \left[ {\begin{array}{*{20}{c}}2\\6\\{ - 1}\end{array}} \right]\).

Short answer

The coordinate vector of \({\mathop{\rm p}\nolimits} \left( t \right) = 1 + 4t + 7{t^2}\) relative to \(B\) is \({\left[ {\mathop{\rm p}\nolimits} \right]_B} = \left[ {\begin{array}{*{20}{c}}2\\6\\{ - 1}\end{array}} \right]\).

Step-by-step solution

Equate the coefficients of the two polynomials

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

\[{c_1}\left( {1 + {t^2}} \right) + {c_2}\left( {t + {t^2}} \right) + {c_3}\left( {1 + 2t + {t^2}} \right) = {\mathop{\rm p}\nolimits} \left( t \right) = 1 + 4t + 7{t^2}\]

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

\[\begin{array}{c}{c_1} + \,\,\,\,\,\,\,{c_3} = 1\\\,\,\,\,\,\,\,\,{c_2} + 2{c_3} = 4\\{c_1} + {c_2} + {c_3} = 7\end{array}\]

Convert the system of equations into an augmented matrix

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

\(\left[ {\begin{array}{*{20}{c}}1&0&1&1\\0&1&2&4\\1&1&1&7\end{array}} \right]\)

Apply the row operation

At row 3, subtract row 1 from row 3.

\( \sim \left[ {\begin{array}{*{20}{c}}1&0&1&1\\0&1&2&4\\0&1&0&6\end{array}} \right]\)

At row 3, subtract row 2 from row 3.

\( \sim \left[ {\begin{array}{*{20}{c}}1&0&1&1\\0&1&2&4\\0&0&{ - 2}&2\end{array}} \right]\)

At row 3, multiply row 3 by \( - \frac{1}{2}\).

\( \sim \left[ {\begin{array}{*{20}{c}}1&0&1&1\\0&1&2&4\\0&0&1&{ - 1}\end{array}} \right]\)

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

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

Determine the coordinate vector of \({\mathop{\rm p}\nolimits} \left( t \right) = 1 + 4t + 7{t^2}\)

Therefore, \({\left[ {\mathop{\rm p}\nolimits} \right]_B} = \left[ {\begin{array}{*{20}{c}}2\\6\\{ - 1}\end{array}} \right]\).

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 \({\mathop{\rm p}\nolimits} \left( t \right) = 1 + 4t + 7{t^2}\) relative to \(B\) is \({\left[ {\mathop{\rm p}\nolimits} \right]_B} = \left[ {\begin{array}{*{20}{c}}2\\6\\{ - 1}\end{array}} \right]\).