# Find a Basis of the Subspace Spanned by Four Polynomials of Degree 3 or Less

## Problem 607

Let $\calP_3$ be the vector space of all polynomials of degree $3$ or less.

Let

\[S=\{p_1(x), p_2(x), p_3(x), p_4(x)\},\]
where

\begin{align*}

p_1(x)&=1+3x+2x^2-x^3 & p_2(x)&=x+x^3\\

p_3(x)&=x+x^2-x^3 & p_4(x)&=3+8x+8x^3.

\end{align*}

**(a)** Find a basis $Q$ of the span $\Span(S)$ consisting of polynomials in $S$.

**(b)** For each polynomial in $S$ that is not in $Q$, find the coordinate vector with respect to the basis $Q$.

*(The Ohio State University, Linear Algebra Midterm)*

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## Solution.

### (a) Find a basis $Q$ of the span $\Span(S)$ consisting of polynomials in $S$.

Let $B=\{1, x, x^2, x^3\}$ be the standard basis for $\calP_3$.

With respect to the basis $B$, the coordinate vectors of the given polynomials are

\begin{align*}

\mathbf{v}_1:=[p_1(x)]_B&=\begin{bmatrix}

1 \\

3 \\

2 \\

-1

\end{bmatrix}, &\mathbf{v}_2:=[p_2(x)]_B=\begin{bmatrix}

0 \\

1 \\

0 \\

1

\end{bmatrix}\\[6pt]
\mathbf{v}_3:=[p_3(x)]_B&=\begin{bmatrix}

0 \\

1 \\

1 \\

-1

\end{bmatrix}, &\mathbf{v}_4:=[p_4(x)]_B=\begin{bmatrix}

3 \\

8 \\

0 \\

8

\end{bmatrix}.

\end{align*}

Let $T=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4\}$ be the set of these coordinate vectors.

We find a basis for $\Span(T)$ among the vectors in $T$ by the leading 1 method.

We form the matrix whose column vectors are the vectors in $T$ and apply elementary row operations as follows.

\begin{align*}

\begin{bmatrix}

1 & 0 & 0 & 3 \\

3 &1 & 1 & 8 \\

2 & 0 & 1 & 0 \\

-1 & 1 & -1 & 8

\end{bmatrix}

\xrightarrow{\substack{R_2-3R_1\\R_3-2R_1\\R_4+R_1}}

\begin{bmatrix}

1 & 0 & 0 & 3 \\

0 &1 & 1 & -1 \\

0 & 0 & 1 & -6 \\

0 & 1 & -1 & 11

\end{bmatrix}

\xrightarrow{R_4-R_2}\\[6pt]
\begin{bmatrix}

1 & 0 & 0 & 3 \\

0 &1 & 1 & -1 \\

0 & 0 & 1 & -6 \\

0 & 0 & -2 & 12

\end{bmatrix}

\xrightarrow{R_4+2R_2}

\begin{bmatrix}

1 & 0 & 0 & 3 \\

0 &1 & 0 & 5 \\

0 & 0 & 1 & -6 \\

0 & 0 & 0 & 0

\end{bmatrix}.

\end{align*}

The first three columns of the reduced row echelon form matrix contain the leading 1’s.

Thus, $\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is a basis for $\Span(T)$.

It follows that $Q:=\{p_1(x), p_2(x), p_3(x)\}$ is a basis for $\Span(S)$.

### (b) For each polynomial in $S$ that is not in $Q$, find the coordinate vector with respect to the basis $Q$.

Note that $p_4(x)$ is not in the basis $Q$.

The fourth column of the matrix in reduced row echelon form of part (a) gives the coefficients of the linear combination:

\[p_4(x)=3p_1(x)+5p_2(x)-6p_3(x).\]

Thus, the coordinate vector of $p_4(x)$ with respect to the basis $Q$ is

\[[p_4(x)]_Q=\begin{bmatrix}

3 \\

5 \\

-6

\end{bmatrix}.\]

## Comment.

This is one of the midterm 2 exam problems for Linear Algebra (Math 2568) in Autumn 2017.

In part (b), some students stopped after obtaining the linear combination $p_4(x)=3p_1(x)+5p_2(x)-6p_3(x)$.

You must read the problem carefully. You are asked to find the coordinate vector of $p_4(x)$ with respect to $Q$.

## List of Midterm 2 Problems for Linear Algebra (Math 2568) in Autumn 2017

- Vector Space of 2 by 2 Traceless Matrices
- Find an Orthonormal Basis of the Given Two Dimensional Vector Space
- Are the Trigonometric Functions $\sin^2(x)$ and $\cos^2(x)$ Linearly Independent?
- Find Bases for the Null Space, Range, and the Row Space of a $5\times 4$ Matrix
- Matrix Representation, Rank, and Nullity of a Linear Transformation $T:\R^2\to \R^3$
- Determine the Dimension of a Mysterious Vector Space From Coordinate Vectors
- Find a Basis of the Subspace Spanned by Four Polynomials of Degree 3 or Less ←The current problem

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