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

Linear algebra problems and solutions

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.

 
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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 $P_2$.


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