Find a basis for $w=\{f\in P_4 : f(1)=f(3)=0\}$.

linear algebravector-spaces

so i am a newbie in linear-algebra and don't know how to find a basis in subspace. the question is find a basis for $w=\{f\in P_4 : f(1)=f(3)=0\}$ where $P_4$ is Polynomial with the maximum power of 4, in other words it is: $a_0 + a_1x^1 + a_2x^2 + a_3x^3 + a_4x^4$.

please walk me trough the steps to find basis and proof that the found basis is for $P_4$.

Best Answer

If $p(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4$, then$$p(1)=a_0+a_1+a_2+a_3+a_4\text{, and }p(3)=a_0+3a_1+9a_2+27a_3+81a_4.$$But\begin{multline}\left\{\begin{array}{l}a_0+a_1+a_2+a_3+a_4=0\\a_0+3a_1+9a_2+27a_3+81a_4=0\end{array}\right.\iff\\\iff a_3=\frac1{27}(-40a_0-39a_1-36a_2)\text{ and }a_4=\frac1{27}(13a_0+12a_1+9a_2).\end{multline}So, you will get a basis if you consider these $3$ polynomials: the one with $a_0=1$, $a_1=0$, and $a_2=0$, the one with $a_0=0$, $a_1=1$, and $a_2=0$, and the one with $a_0=0$, $a_1=0$ and $a_2=1$. For instance, the first of these will be$$1-\frac{40}{27}x^3+\frac{13}{27}x^4.$$

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