# Basis and Dimension of the Subspace of All Polynomials of Degree 4 or Less Satisfying Some Conditions.

## Problem 256

Let $P_4$ be the vector space consisting of all polynomials of degree $4$ or less with real number coefficients.
Let $W$ be the subspace of $P_2$ by
$W=\{ p(x)\in P_4 \mid p(1)+p(-1)=0 \text{ and } p(2)+p(-2)=0 \}.$ Find a basis of the subspace $W$ and determine the dimension of $W$.

## Proof.

Let $p(x)\in W$ and write
$p(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4$ for some coefficients $a_0, a_1, a_2, a_3, a_4 \in \R$.

We first compute the values of $p(x)$ at $x=\pm 1$ and $x=\pm 2$.
We have
\begin{align*}
p(1)&=a_0+a_1+a_2+a_3+a_4\\
p(1)&=a_0-a_1+a_2-a_3+a_4\\
p(2)&=a_0+2a_1+4a_2+8a_3+16a_4\\
p(-2)&=a_0-2a_1+4a_2-8a_3+16a_4.
\end{align*}

Since $p(x)$ is in $W$, it satisfies
$p(1)+p(-1)=0 \text{ and } p(2)+p(-2)=0,$ and thus we have
\begin{align*}
p(1)+p(-1)&=2a_0+2a_2+2a_4=0\\
p(2)+p(-1)&=2a_0+8a_2+32a_4=0.\\
\end{align*}
Dividing them by $2$, we have the system of linear equations in $a_0, a_2, a_4$.
\begin{align*}
a_0+a_2+a_4&=0\\
a_0+4a_2+16a_4&=0.
\end{align*}

We reduce the augmented matrix of the system by elementary row operations as follows.
\begin{align*}
\left[\begin{array}{rrr|r}
1 & 1 & 1 & 0 \\
1 & 4 & 16 & 0
\end{array} \right] \xrightarrow{R_2-R_1}
\left[\begin{array}{rrr|r}
1 & 1 & 1 & 0 \\
0 & 3 & 15 & 0
\end{array} \right] \xrightarrow{\frac{1}{3}R_2}\\
\left[\begin{array}{rrr|r}
1 & 1 & 1 & 0 \\
0 & 1 & 5 & 0
\end{array} \right] \xrightarrow{R_1-R_2}
\left[\begin{array}{rrr|r}
1 & 0 & -4 & 0 \\
0 & 1 & 5 & 0
\end{array} \right].
\end{align*}

Hence the solutions to the system is
\begin{align*}
a_0&=4a_4\\
a_2&=-5a_4
\end{align*}
and $a_4$ is a free variable.

Substituting these relations into $p(x)$, we obtain
\begin{align*}
p(x)&=4a_4+a_1x-5a_4x^2+a_3x^3+a_4x^4\\
&=a_1x+a_3x^3+a_4(4-5x^2+x^4).
\end{align*}

Let
\begin{align*}
p_1(x)&=x\\
p_2(x)&=x^3\\
p_3(x)&=4-5x^2+x^4.
\end{align*}
These are vectors in $W$.
(To see this, you may directly check the defining relations, or set $a_1=a_3=0$ and $a_4=1$ to get $p_3(x)$. Similarly for $p_1(x)$ and $p_2(x)$.)

Then by the above computations, any vector $p(x)$ in $W$ is a linear combination
$p(x)=a_1p_1(x)+a_3p_2(x)+a_4p_3(x).$ Thus, $\{p_1(x), p_2(x), p_3(x)\}$ is a spanning set of $W$.

Also, the vectors $p_1(x), p_2(x), p_3(x)$ are linearly independent.
In fact, if we have $c_1p_1(x)+c_2p_2(x)+c_3p_3(x)=0$, then we have
\begin{align*}
0&=c_1p_1(x)+c_2p_2(x)+c_3p_3(x)\\
&=c_1x+c_2x^3+c_3(4-5x^2+x^4)\\
&=4c_3+c_1x-5c_3x^2+c_2x^3+c_3x^4.
\end{align*}

Since the coefficients must be zero, we have $c_1=c_2=c_3=0$.
This proves that $p_1(x), p_2(x), p_3(x)$ are linearly independent.

Therefore the set $\{p_1(x), p_2(x), p_3(x)\}$ is linearly independent spanning set of the subspace $W$, hence it is a basis for $W$.
Thus the dimension of the subspace $W$ is $3$.

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