Linear dependency of vectors in the dual space

Question

Q : Find a basis for $\textsf W^0$ (the annihilator of $\textsf W$) and find $\dim(\textsf W^0)$, when $$\textsf W = \operatorname{span}(\{\phi_0,\phi_1\}) \subseteq (\mathbb C_{\le 3}[x])^*$$ and $\phi_a(f)=f(a)$ for all $f\in (\mathbb C_{\le 3}[x])^*$.

So here's what i'm struglling with:

• $\textsf W^0 = \big( \operatorname{span}(\{\phi_0,\phi_1\}) \big)^0 = (\{\phi_0,\phi_1\})^0 = \{v\in\mathbb C_{\le3}[x] :\, \phi_0(v)=\phi_1(v)=0\}$.
• $\dim(\textsf W^0) = \dim(\mathbb C_{\le 3}[x]) - \dim(\textsf W)$.

So I need to know the linear dependency of $\varphi_0,\varphi_1$ to conclude how many vectors spans $\textsf W^0$.
Now I can express $\phi_0,\phi_1$ as :

• $\phi_0(a+bx+cx^2+dx^3)=a\cdot\phi_0(1)+b\cdot\phi_0(x)+c\cdot\phi_0(x^2)+d\cdot \phi_0(x^3)=a$
• $\phi_1(a+bx+cx^2+dx^3)=a+b+c+d$

since $\phi_a(f)=f(a)$. (Plese let me know if this use is right, because $\phi_0(1)=1(0)$ feels weird to me, but makes sense).

From here, how do I find the linear dependency?

Answer 1

Consider a more general situation. Suppose $V$ is a finite dimensional vector space and that $\alpha_1,\dots,\alpha_n\in V^*$.

If $W=\operatorname{span}\{\alpha_1,\dots,\alpha_n\}$, then $$ W^0=\{v\in V: \alpha_k(v)=0,\quad i=1,2,\dots,n\} $$ (prove it). There's no need to check whether $\{\alpha_1,\dots,\alpha_n\}$ is linearly independent or not.

In your case, $$ W^0=\{f\in\mathbb{C}_{\le3}[x]:\phi_0(f)=0,\phi_1(f)=0\} $$ If $f(x)=a_0+a_1x+a_2x^2+a_3x^3$, the conditions read $$ \begin{cases} a_0=0 \\[5px] a_0+a_1+a_2+a_3=0 \end{cases} $$ By the way, since $\dim W^0=2$, you can conclude that $\dim W=4-2=2$, so $\{\phi_0,\phi_1\}$ is linearly independent.

A basis for $W^0$ is $\{x^2-x,x^3-x\}$.

Answer 2

Here is my answer. Regarding the linear independence. Let's just verify the definition :
Our set $W_0$ is lin $\iff$ ($\sum_{i=0}^{1} \alpha_i \, \phi_i = 0 (\alpha_i \text{ being polynomials}) \implies \alpha_i = 0 )$
As you said, we indeed have : $\phi_0(f)= a$ and $\phi_1(f)= a + b + c + d$
By injecting this into the lin formula : $$ (\alpha_0+\alpha_1)a + \alpha_1(b+c+d) = 0 \\ $$ $\alpha_1(b+c+d) = 0$ is only satisfied for $\alpha_1 = 0$ (since all $a,b,c,d$ are in $\mathbb{C}$ and can take any value) $\implies$ $\alpha_0a = 0 \, \forall a\in \mathbb{C}$ which in turn implies $\alpha_0$ is the null polynomial. That is it for linear independency.