# A Basis for the Vector Space of Polynomials of Degree Two or Less and Coordinate Vectors

## Problem 150

Show that the set
$S=\{1, 1-x, 3+4x+x^2\}$ is a basis of the vector space $P_2$ of all polynomials of degree $2$ or less.

## Proof.

We know that the set $B=\{1, x, x^2\}$ is a basis for the vector space $P_2$.
With respect to this basis $B$, the coordinate vectors of vectors in $S$ are
$[1]_B=\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \quad [1-x]_B=\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}, \quad [3+4x+x^2]_B=\begin{bmatrix} 3 \\ 4 \\ 1 \end{bmatrix}.$

Recall the fact that the set $S$ is a basis for $P_2$ if and only if the coordinate vectors
$T=\{[1]_B, [1-x]_B, [3+4x+x^2]_B\}$ is a basis for $R^3$.

Hence we check that $T$ is a basis for $R^3$. Note that $T$ is a linearly independent set.
In fact, if we have
$a_1 \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} +a_2 \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} +a_3 \begin{bmatrix} 3 \\ 4 \\ 1 \end{bmatrix}=\mathbf{0},$ then we have
$\begin{bmatrix} a_1+a_2+3a_3 \\ -a_2+4a_3 \\ a_3 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$ and we obtain $a_1=a_2=a_3=0$, and thus $T$ is a linearly independent set.

Since $T$ consists of three linearly independent vectors in $\R^3$ and the dimension of $\R^3$ is $3$, the set $T$ must be a basis for $\R^3$.
Therefore, the set $S$ is a basis for $P_2$.

### More from my site

#### You may also like...

##### Nilpotent Matrices and Non-Singularity of Such Matrices

Let $A$ be an $n \times n$ nilpotent matrix, that is, $A^m=O$ for some positive integer $m$, where $O$ is...

Close