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

Linear Algebra Problems and Solutions

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 $W$.

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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

Since $p(x)$ is in $W$, it satisfies
\[p(1)+p(-1)=0 \text{ and } p(2)+p(-2)=0,\] and thus we have
Dividing them by $2$, we have the system of linear equations in $a_0, a_2, a_4$.

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

Hence the solutions to the system is
and $a_4$ is a free variable.

Substituting these relations into $p(x)$, we obtain

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

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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