# Basis of Span in Vector Space of Polynomials of Degree 2 or Less ## Problem 367

Let $P_2$ be the vector space of all polynomials of degree $2$ or less with real coefficients.
Let
$S=\{1+x+2x^2, \quad x+2x^2, \quad -1, \quad x^2\}$ be the set of four vectors in $P_2$.

Then find a basis of the subspace $\Span(S)$ among the vectors in $S$.

(Linear Algebra Exam Problem, the Ohio State University) Add to solve later

## Solution.

Let $B=\{1, x, x^2\}$ be the standard basis of the vector space $P_2$.
With respect to the basis $B$, the coordinate vectors of vectors in $S$ are
$[1+x+2x^2]_B=\begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}, \quad [x+2x^2]_B=\begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}, \quad [-1]_B=\begin{bmatrix} -1 \\ 0 \\ 0 \end{bmatrix}, \quad [x^2]_B=\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}.$ Let
$T=\left\{\, [1+x+2x^2]_B, [x+2x^2]_B, [-1]_B, [x^2]_B \,\right\}$ be the set of these coordinate vectors.

We then find a basis of $\Span(T)$ among vectors in $T$ by the leading 1 method.
We reduce the augmented matrix by elementary row operations as follows. We have
\begin{align*}
\begin{bmatrix}
1 & 0 & -1 & 0 \\
1 &1 & 0 & 0 \\
2 & 2 & 0 & 1
\end{bmatrix}
\xrightarrow{R_3-2R_2}
\begin{bmatrix}
1 & 0 & -1 & 0 \\
1 &1 & 0 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}
\xrightarrow{R_2-R_1}
\begin{bmatrix}
1 & 0 & -1 & 0 \\
0 &1 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}.
\end{align*}

The last matrix is in reduced row echelon form and the first, the second, and the fourth columns contain the leading 1’s.
Therefore, it follows that
$\left\{\, [1+x+2x^2]_B, [x+2x^2]_B, [x^2]_B \,\right\}$ is a basis of $\Span(T)$, and hence
$\{1+x+2x^2, \quad x+2x^2, \quad x^2\}$ is a basis of $\Span(S)$ consisting of the vectors of $S$.

## Linear Algebra Midterm Exam 2 Problems and Solutions Add to solve later

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