Finding the basis for a subspace of P3 How would I find a basis for the set $V \subset P_{3} \subset C[-1,1]$ defined as all $f \in P_{3}$ with $f(-1)=0$ and $\int_{-1}^{1}f(x)dx = 0$? Should I be looking for a specific polynomial that satisfies both of the conditions of $V$?
Yes, you should. We know that every $f\in P_3$ is of the form $a+bx+cx^2+dx^3$; then $f(-1)=0$ implies that $a-b+c-d=0$.
Then $\int_{-1}^1f(x)dx=ax+\frac b2x^2+\frac c3 x^3+\frac d4x^4\Bigg|_{-1}^1=2a+\frac{2c}3=0$
We can simplify these down by finding the null space of the corresponding matrix $\begin{bmatrix}1&amp;-1&amp;1&amp;-1\\2&amp;0&amp;\frac23&amp;0\end{bmatrix}\sim\begin{bmatrix}1&amp;0&amp;\frac13&amp;0\\0&amp;1&amp;-\frac23&amp;1\end{bmatrix}$, and so the null space is spanned by $\begin{bmatrix}1\\0\\-3\\-2\end{bmatrix}$ and $\begin{bmatrix}0\\1\\0\\-1\end{bmatrix}$.
Therefore, a basis for $V$ is the polynomials $1-3x^2-2x^3$ and $x-x^3$.
Yes, you should. We know that every $f\in P_3$ is of the form $a+bx+cx^2+dx^3$; then $f(-1)=0$ implies that $a-b+c-d=0$.
Then $\int_{-1}^1f(x)dx=ax+\frac b2x^2+\frac c3 x^3+\frac d4x^4\Bigg|_{-1}^1=2a+\frac{2c}3=0$
We can simplify these down by finding the null space of the corresponding matrix $\begin{bmatrix}1&amp;-1&amp;1&amp;-1\\2&amp;0&amp;\frac23&amp;0\end{bmatrix}\sim\begin{bmatrix}1&amp;0&amp;\frac13&amp;0\\0&amp;1&amp;-\frac23&amp;1\end{bmatrix}$, and so the null space is spanned by $\begin{bmatrix}1\\0\\-3\\-2\end{bmatrix}$ and $\begin{bmatrix}0\\1\\0\\-1\end{bmatrix}$.
Therefore, a basis for $V$ is the polynomials $1-3x^2-2x^3$ and $x-x^3$.