# 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 $S$. Add to solve later

## 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
$[p(x)]_S=\begin{bmatrix} 19 \\ -8 \\ 10 \end{bmatrix}.$ Add to solve later

### More from my site

#### You may also like...

This site uses Akismet to reduce spam. Learn how your comment data is processed.

###### More in Linear Algebra ##### Commuting Matrices $AB=BA$ such that $A-B$ is Nilpotent Have the Same Eigenvalues

Let $A$ and $B$ be square matrices such that they commute each other: $AB=BA$. Assume that $A-B$ is a nilpotent...

Close