# Find a Basis of the Vector Space of Polynomials of Degree 2 or Less Among Given Polynomials

## Problem 481

Let $P_2$ be the vector space of all polynomials with real coefficients of degree $2$ or less.
Let $S=\{p_1(x), p_2(x), p_3(x), p_4(x)\}$, where
\begin{align*}
p_1(x)&=-1+x+2x^2, \quad p_2(x)=x+3x^2\\
p_3(x)&=1+2x+8x^2, \quad p_4(x)=1+x+x^2.
\end{align*}

(a) Find a basis of $P_2$ among the vectors of $S$. (Explain why it is a basis of $P_2$.)

(b) Let $B’$ be the basis you obtained in part (a).
For each vector of $S$ which is not in $B’$, find the coordinate vector of it with respect to the basis $B’$.

(The Ohio State University, Linear Algebra Final Exam Problem)

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## Solution.

### (a) Find a basis of $P_2$ among the vectors of $S$.

The vector space $P_2$ has a basis $B=\{1, x, x^2\}$ and the dimension of $P_2$ is $3$.

The coordinate vectors with respect to this basis are
\begin{align*}
[p_1(x)]_B&=\begin{bmatrix}
-1 \\
1 \\
2
\end{bmatrix}, \quad [p_2(x)]_B=\begin{bmatrix}
0 \\
1 \\
3
\end{bmatrix}\6pt] [p_3(x)]_B&=\begin{bmatrix} 1 \\ 2 \\ 8 \end{bmatrix}, \quad [p_4(x)]_B=\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}. \end{align*} Consider the vector space \Span(T) spanned by \[T=\{[p_1(x)]_B, [p_2(x)]_B, [p_3(x)]_B, [p_4(x)]_B \}.

We determine a basis vectors of $\Span(T)$ among vectors in $T$.
Consider the matrix $A$ whose column vectors are these coordinate vectors:
$A=\begin{bmatrix} -1 & 0 & 1 & 1 \\ 1 &1 & 2 & 1 \\ 2 & 3 & 8 & 1 \end{bmatrix}.$ Applying the elementary row operations, we obtain
\begin{align*}
A&=\begin{bmatrix}
-1 & 0 & 1 & 1 \\
1 &1 & 2 & 1 \\
2 & 3 & 8 & 1
\end{bmatrix}
\xrightarrow{\substack{R_2+R_1\\ R_3+2R_1}}
\begin{bmatrix}
-1 & 0 & 1 & 1 \\
0 &1 & 3 & 2 \\
0 & 3 & 10 & 3
\end{bmatrix}\6pt] & \xrightarrow[R_3-3R_2]{-R_1} \begin{bmatrix} 1 & 0 & -1 & -1 \\ 0 &1 & 3 & 2 \\ 0 & 0 & 1 & -3 \end{bmatrix} \xrightarrow{\substack{R_1+R_3\\ R_2-3R_3}} \begin{bmatrix} 1 & 0 & 0 & -4 \\ 0 &1 & 0 & 11 \\ 0 & 0 & 1 & -3 \end{bmatrix}. \end{align*} The first three columns contain the leading 1’s. By the leading 1 method, we know that \[\{[p_1(x)]_B, [p_2(x)]_B, [p_3(x)]_B\} is a basis of the vector space $\Span(T)$.

It follows by the correspondence theorem that
$\{p_1(x), p_2(x), p_3(x)\}$ is a basis of $\Span(S)$.

This yields that $\Span(S)$ is a three dimensional subspace of $P_2$, which is also three dimensional.
Hence we have $P_2=\Span(S)$.
Therefore
$\{p_1(x), p_2(x), p_3(x)\}$ is a basis of $P_2$.

### (b) For each vector of $S$ which is not in $B’$, find the coordinate vector of it with respect to the basis $B’$.

In part (a), we found a basis
$B’=\{p_1(x), p_2(x), p_3(x)\}$ of the vector space $P_2$.
So, we need to find the coordinate vector of $p_4(x)$ with respect to $B’$.

Thus, we need to solve
$ap_1(x)+bp_2(x)+cp_3(x)=p_4(x)$ for $a, b, c$.

Using the coordinate vectors, this is equivalent to solve
$\begin{bmatrix} -1 & 0 & 1 \\ 1 &1 & 2 \\ 2 & 3 & 8 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix}=\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}.$ The augmented matrix of this system is exactly the matrix $A$ we dealt in part (a).
By the reduction of $A$ obtained in part (a) yields that
$a=-4, b=11, c=-3.$

Hence we obtain
$p_4(x)=-4p_1(x)+11p_2(x)+3p_3(x),$ and thus the coordinate vector of $p_4(x)$ with respect to $B’$ is
$[p_4(x)]_{B’}=\begin{bmatrix} -4 \\ 11 \\ 3 \end{bmatrix}.$

Remark that this is the last column vector of the reduced row echelon form matrix of $A$.

## Final Exam Problems and Solution. (Linear Algebra Math 2568 at the Ohio State University)

This problem is one of the final exam problems of Linear Algebra course at the Ohio State University (Math 2568).

The other problems can be found from the links below.

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1. 06/28/2017

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