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

## 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
\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 .

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 $B$:
$[ -3 + 2x^3 ]_{B} = \begin{bmatrix} -2 \\ 0 \\ 1 \\ -1 \end{bmatrix}.$

Let $V$ denote the vector space of $2 \times 2$ matrices, and $W$ the vector space of $3 \times 2$...