The Coordinate Vector for a Polynomial with respect to the Given Basis

Vector Space Problems and Solutions

Problem 683

Let $\mathrm{P}_3$ denote the set of polynomials of degree $3$ or less with real coefficients. Consider the ordered basis
\[B = \left\{ 1+x , 1+x^2 , x – x^2 + 2x^3 , 1 – x – x^2 \right\}.\] Write the coordinate vector for the polynomial $f(x) = -3 + 2x^3$ in terms of the basis $B$.

 
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Solution.

We will use the standard basis $C = \{ 1 , x , x^2 , x^3 \}$ to write the coordinate vectors for $f$ and each element of $B$. We then use the simpler coordinate vectors to find a solution to the equation
\begin{equation}\tag{*}
c_1 ( 1 + x ) + c_2 ( 1 + x^2 ) + c_3 ( x – x^2 + 2 x^3 ) + c_4 ( 1 – x – x^2 ) = -3 + 2 x^3 . \end{equation}

The coordinate vectors are found by writing an element as a linear sum of elements of $C$. The coefficients in the linear sum become the entries in the coordinate vector. Thus we have the following coordinate vectors:
\[[ 1 + x ]_{C} = \begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix} , \, [ 1 + x^2 ]_{C} = \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix} , \, [ x – x^2 + 2 x^3 ]_{C} = \begin{bmatrix} 0 \\ 1 \\ -1 \\ 2 \end{bmatrix},\] \[[ 1 – x – x^2 ]_{C} = \begin{bmatrix} 1 \\ -1 \\ -1 \\ 0 \end{bmatrix} , \, [ -3 + 2 x^3 ]_{C} = \begin{bmatrix} -3 \\ 0 \\ 0 \\ 2 \end{bmatrix}.\]

Plugging these coordinate vectors into Equation (*), we get the equation
\[c_1 \begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix} + c_3 \begin{bmatrix} 0 \\ 1 \\ -1 \\ 2 \end{bmatrix} + c_4 \begin{bmatrix} 1 \\ -1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 0 \\ 0 \\ 2 \end{bmatrix}.\]


We can now create the augmented matrix which represents this equation:
\[ \left[\begin{array}{rrrr|r}
1 & 1 & 0 & 1 & -3 \\ 1 & 0 & 1 & -1 & 0 \\ 0 & 1 & -1 & -1 & 0 \\ 0 & 0 & 2 & 0 & 2
\end{array} \right].\]


Now the equation is solved by reducing the augmented matrix:
\begin{align*}
\left[\begin{array}{rrrr|r} 1 & 1 & 0 & 1 & -3 \\ 1 & 0 & 1 & -1 & 0 \\ 0 & 1 & -1 & -1 & 0 \\ 0 & 0 & 2 & 0 & 2 \end{array} \right] \xrightarrow{ R_2 – R_1 } \left[\begin{array}{rrrr|r} 1 & 1 & 0 & 1 & -3 \\ 0 & -1 & 1 & -2 & 3 \\ 0 & 1 & -1 & -1 & 0 \\ 0 & 0 & 2 & 0 & 2 \end{array} \right] \xrightarrow{ (-1) R_2 } \left[\begin{array}{rrrr|r} 1 & 1 & 0 & 1 & -3 \\ 0 & 1 & -1 & 2 & -3 \\ 0 & 1 & -1 & -1 & 0 \\ 0 & 0 & 2 & 0 & 2 \end{array} \right] \\[6pt] \xrightarrow[R_3 – R_2]{ R_1 – R_2 } \left[\begin{array}{rrrr|r} 1 & 0 & 1 & -1 & 0 \\ 0 & 1 & -1 & 2 & -3 \\ 0 & 0 & 0 & -3 & 3 \\ 0 & 0 & 2 & 0 & 2 \end{array} \right] \xrightarrow[ \frac{1}{2} R_4]{ \left( \frac{-1}{3} \right) R_3 } \left[\begin{array}{rrrr|r} 1 & 0 & 1 & -1 & 0 \\ 0 & 1 & -1 & 2 & -3 \\ 0 & 0 & 0 & 1 & -1 \\ 0 & 0 & 1 & 0 & 1 \end{array} \right] \\[6pt] \xrightarrow{ R_3 \leftrightarrow R_4 } \left[\begin{array}{rrrr|r} 1 & 0 & 1 & -1 & 0 \\ 0 & 1 & -1 & 2 & -3 \\ 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & -1 \end{array} \right] \xrightarrow[ R_2 – 2 R_4 ]{ R_1 + R_4 } \left[\begin{array}{rrrr|r} 1 & 0 & 1 & 0 & -1 \\ 0 & 1 & -1 & 0 & -1 \\ 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & -1 \end{array} \right] \\[6pt] \xrightarrow[R_2 + R_3]{ R_1 – R_3 } \left[\begin{array}{rrrr|r} 1 & 0 & 0 & 0 & -2 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & -1 \end{array} \right] .
\end{align*}


We can now read off the solution $c_1 = -2$, $c_2 = 0$ , $c_3 = 1$, and $c_4 = -1$. We can now plug these coefficients into Equation (*), yielding the equation
\[-2 (1 + x) + ( x – x^2 + 2x^3 ) – ( 1 – x – x^2 ) = -3 + 2x^3.\]

Finally, we use these coefficients to write the coordinate vector for $f$ in terms of the basis $B$:
\[ [ -3 + 2x^3 ]_{B} = \begin{bmatrix} -2 \\ 0 \\ 1 \\ -1 \end{bmatrix}.\]


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