# Use Coordinate Vectors to Show a Set is a Basis for the Vector Space of Polynomials of Degree 2 or Less

## Problem 588

Let $P_2$ be the vector space over $\R$ of all polynomials of degree $2$ or less.

Let $S=\{p_1(x), p_2(x), p_3(x)\}$, where

\[p_1(x)=x^2+1, \quad p_2(x)=6x^2+x+2, \quad p_3(x)=3x^2+x.\]

**(a)** Use the basis $B=\{x^2, x, 1\}$ of $P_2$ to prove that the set $S$ is a basis for $P_2$.

**(b)** Find the coordinate vector of $p(x)=x^2+2x+3\in P_2$ with respect to the basis $S$.

Contents

## Solution.

### (a) Prove that the set $S$ is a basis for $P_2$.

The coordinate vectors of $p_1(x)$ with respect to the basis $B=\{x^2, x, 1\}$ is given by

\[[p_1(x)]_B=\begin{bmatrix}

1 \\

0 \\

1

\end{bmatrix}\]
as $p_(x)$ can be written as a linear combination $p_1(x)=1\cdot x^2+0\cdot x+1\cdot 1$ of the basis vectors in $B$.

Similarly, we have

\[[p_2(x)]_B=\begin{bmatrix}

6 \\

1 \\

2

\end{bmatrix} \text{ and } [p_3(x)]_B=\begin{bmatrix}

3 \\

1 \\

0

\end{bmatrix}.\]

Let $T=\{[p_1(x)]_B, [p_2(x)]_B, [p_3(x)]_B\}$.

Then $S$ is a basis for $P_2$ if and only if $T$ is a basis for $\R^3$.

Thus it remains to show that $T$ is a basis for $\R^3$.

Consider the matrix whose column vectors are vectors in $T$.

We have

\begin{align*}

\begin{bmatrix}

1 & 6 & 3 \\

0 &1 &1 \\

1 & 2 & 0

\end{bmatrix}

\xrightarrow{R_3-R_1}

\begin{bmatrix}

1 & 6 & 3 \\

0 &1 &1 \\

0 & -4 & -3

\end{bmatrix}

\xrightarrow{\substack{R_1-6R_2\\R_3+4R_2}}\\[6pt] \begin{bmatrix}

1 & 0 & -3 \\

0 &1 &1 \\

0 & 0 & 1

\end{bmatrix}

\xrightarrow{\substack{R_1+3R_3\\R_2-R_3}}

\begin{bmatrix}

1 & 0 & 0 \\

0 &1 &0 \\

0 & 0 & 1

\end{bmatrix}.

\end{align*}

It follows that $T$ is linearly independent.

As $T$ consists of three linearly independent vectors in $\R^3$, it is a basis for $\R^3$.

Hence $S$ is a basis for $P_2$.

### (b) Find the coordinate vector of $p(x)=x^2+2x+3\in P_2$ with respect to the basis $S$.

To find the coordinate vector $[p(x)]_S$ with respect to the basis $S$, we express $p(x)$ as a linear combination of the basis vectors in $S$.

Thus we want to find scalars $c_1, c_2, c_3$ such that

\[p(x)=c_1p_1(x)+c_2p_2(x)+c_3p_3(x).\]
Considering the coordinate vectors of both sides, this is equivalent to

\[\begin{bmatrix}

1 \\

2 \\

3

\end{bmatrix}=c_1\begin{bmatrix}

1 \\

0 \\

1

\end{bmatrix}+c_2\begin{bmatrix}

6 \\

1 \\

2

\end{bmatrix}+c_3\begin{bmatrix}

3 \\

1 \\

0

\end{bmatrix}.\]
We write this equation as a matrix equation

\[\begin{bmatrix}

1 & 6 & 3 \\

0 &1 &1 \\

1 & 2 & 0

\end{bmatrix}\begin{bmatrix}

c_1 \\

c_2 \\

c_3

\end{bmatrix}=\begin{bmatrix}

1 \\

2 \\

3

\end{bmatrix}.\]

To solve this, we apply elementary row operations to the augmented matrix as follows.

\begin{align*}

\left[\begin{array}{rrr|r}

1 & 6 & 3 & 1 \\

0 &1 & 1 & 2 \\

1 & 2 & 0 & 3

\end{array} \right]
\xrightarrow{R_3-R_1}

\left[\begin{array}{rrr|r}

1 & 6 & 3 & 1 \\

0 &1 & 1 & 2 \\

0 & -4 & -3 & 2

\end{array} \right]
\xrightarrow{\substack{R_1-6R_2\\R_3+4R_2}}\\[6pt]
\left[\begin{array}{rrr|r}

1 & 0 & -3 & -11 \\

0 &1 & 1 & 2 \\

0 & 0 & 1 & 10

\end{array} \right]
\xrightarrow{\substack{R_1+3R_3\\R_2-R_3}}

\left[\begin{array}{rrr|r}

1 & 0 & 0 & 19 \\

0 &1 & 0 & -8 \\

0 & 0 & 1 & 10

\end{array} \right].

\end{align*}

(Note that elementary row operations are exactly the same as before.)

Hence the solution is $c_1=19, c_2=-8, c_3=10$.

Thus, we have the linear combination

\[p(x)=19p_1(x)-8p_2(x)+10p_3(x)\]
and the coordinate vector of $p(x)$ with respect to the basis $S$ is

\[[p(x)]_S=\begin{bmatrix}

19 \\

-8 \\

10

\end{bmatrix}.\]

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