Let $P_3$ be the vector space over $\R$ of all degree three or less polynomial with real number coefficient.
Let $W$ be the following subset of $P_3$.
\[W=\{p(x) \in P_3 \mid p'(-1)=0 \text{ and } p^{\prime\prime}(1)=0\}.\] Here $p'(x)$ is the first derivative of $p(x)$ and $p^{\prime\prime}(x)$ is the second derivative of $p(x)$.

Show that $W$ is a subspace of $P_3$ and find a basis for $W$.

Proof.

Subspace criteria

To show that the subset $W$ of the vector space $P_3$ is a subspace, we need to check that

Check Condition 1

First note that the zero vector in $P_3$ is the zero polynomial, which we denote $\theta(x)$.
Thus we have $\theta(x)=0$ for any $x$.

The derivative of the zero function is still the zero function, we have
\[\theta'(x)=0, \text{ and } \theta^{\prime\prime}(x)=0.\]

Therefore the zero vector $\theta(x)$ satisfies the following defining relations for $W$
\[\theta'(-1)=0 \text{ and } \theta^{\prime\prime}(1)=0\] and thus the zero vector $\theta(x)$ is in $W$.
Thus condition 1 is met.

Check Condition 2

To check condition 2, let $f(x), g(x) \in W$.
Then $f(x)$ and $g(x)$ satisfy
\[f'(-1)=0, f^{\prime\prime}(1)=0, \text{ and } g'(-1)=0, g^{\prime\prime}(1)=0.\] Let $h(x):=f(x)+g(x)$. We want to show that the sum $h(x)=f(x)+g(x)$ is in $W$.

We have
\[h'(x)=(f(x)+g(x))’=f'(x)+g'(x)\] and similarly we have
\[h^{\prime\prime}(x)=f^{\prime\prime}(x)+g^{\prime\prime}(x).\]

Thus we obtain
\[h'(-1)=f'(-1)+g'(-1)=0+0=0 \text{ and } h^{\prime\prime}(1)=f^{\prime\prime}(1)+g^{\prime\prime}(1)=0+0=0.\] Therefore $h(x)$ satisfies the defining relations of $W$ and hence $h(x)=f(x)+g(x) \in W$.
So condition 2 is satisfied.

Check Condition 3

Condition 3 can be checked as follows. Let $f(x) \in W$ and let $r\in R$.
We want to show that the scalar product $k(x):=rf(x)$ is in $W$.

Since $f(x)$ is in $W$, we have
\[f'(-1)=0, f^{\prime\prime}(1)=0.\]

Note that we have
\[k'(x)=(rf(x))’=rf'(x), \text{ and } k^{\prime\prime}(x)=rf^{\prime\prime}(x).\]

Thus we see that
\[k'(-1)=rf'(-1)=r\cdot 0=0 \text{ and } k^{\prime\prime}(1)=rf^{\prime\prime}(1)=r\cdot 0=0.\]

Thus $k(x)$ satisfies the defining relations of $W$ and hence the scalar product $k(x)=rf(x)$ is in $W$.
Therefore condition 3 and thus all conditions are met, and we conclude that $W$ is a subspace of $P_3$.

Find a basis for the subspace $W$.

To find a basis, we observe the following.

Any vector in $W$ is a polynomial
\[p(x)=a_0+a_1x+a_2x^2+a_3x^3\] satisfying
\[p'(-1)=0 \text{ and } p^{\prime\prime}(1)=0.\]

Since we have
\[p'(x)=a_1+2a_2x+3a_3x^2 \text{ and } p^{\prime\prime}(x)=2a_2+6a_3x,\] the above conditions become
\[a_1-2a_2+3a_3=0 \text{ and } 2a_2+6a_3=0.\] To find solutions for these equations consider the augmented matrix
\[\left[\begin{array}{rrr|r}
1 & -2 & 3 & 0 \\
0 &2 & 6 & 0
\end{array} \right].\] We apply the elementary row operations to this matrix and obtain the following reduced row echelon matrix.
\[\left[\begin{array}{rrr|r}
1 & 0 & 9 & 0 \\
0 &1 & 3 & 0
\end{array} \right].\] Thus, solutions are\begin{align*}
a_1&=-9a_3\\
a_2&=-3a_3.
\end{align*}

Therefore, any polynomial $p(x)$ in $W$ can be written as
\begin{align*}
p(x)&=a_0-9a_3x-3a_3x^2+a_3x^3\\
&=a_0(1)+a_3(-9x-3x^2+x^3).
\end{align*}

In particular we see that the polynomials $q_1(x):=1$ and $q_2(x):=-9x-3x^2+x^3$ are in $W$.
Since any vector $p(x) \in W$ is a linear combination of $q_1(x)$ and $q_2(x)$, the set $\{q_1(x), q_2(x)\}$ is a spanning set for $W$.

We check that $q_1(x)$ and $q_2(x)$ are linearly independent.
If we have a linear combination
\[c_1q_1(x)+c_2q_2(x)=0(=\theta(x)),\] then we have
\[c_1-9c_2x-3c_2x^2+c_2x^3=0.\]

Thus we see that $c_1=c_2=0$ and the vectors $q_1(x)$ and $q_2(x)$ are linearly independent.
Therefore $\{q_1(x), q_2(x)\}$ is a linearly independent spanning set for $W$, and thus it is a basis for $W$.

In summary we found a basis
\[\{1, -9x-3x^2+x^3\}.\]

Remark: The dimension of $W$ is $2$.

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