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 $S$.

 
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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
\[[p(x)]_S=\begin{bmatrix}
19 \\
-8 \\
10
\end{bmatrix}.\] Click here if solved 28

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