# 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) Add to solve later

## 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 Add to solve later

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