# Proving that a set of functionals mapping $P_1(x)$ to the reals is a basis for the dual space of $P_1(x)$

Given a vector space $V = P_1(\mathbb{R})$ and $f_1,f_2 \in V^*$ where $V^*$ is the dual space of $V$ and $$f_1 = \int_0^1 p(t) \, dt \quad\text{and}\quad f_2 = \int_0^2 p(t) \, dt$$ prove that $\{f_1,f_2\}$ is a basis for $V^*$ and find a basis for $V$ for which it is the dual basis.

I tried using the classical argument for a basis.

Let $p(x) = a + bx$.

By a theorem, we know that $\dim(V) = \dim(V^*)$. Since we have 2 vectors and $\dim(V) = 2$ it remains to show that $\{f_1,f_2\}$ is linearly independent in order for it to be a basis. In order for $\{f_1,f_2\}$ to be linearly independent, it must satisfy that for any $v = p(x) \in V$, $$c_1f_1(v) + c_2f_2(v) = 0 \implies c_1 = c_2 = 0$$

I get as far as: $$c_1(a+\frac{b}{2}) + c_2(2a + 2b) = 0$$

However, since $a,b$ are both elements of $\mathbb{R}$, the system is underdetermined and I have an infinite number of solutions. i.e. $$c_1(a+\frac{b}{2}) = c_2(-2a - 2b)$$

Wouldn't this suggest that $\{f_1,f_2\}$ is not linearly independent?

The solution manual I am looking at only offers the following: We know that \begin{align*} f_1(p(x)) &= f_1(a+bx) = a+\frac{b}{2} \\ f_2(p(x)) &= f_2(a+bx) = 2a+2b \end{align*}

Then, it goes on to say that: $$a+bx = (a+\frac{b}{2})(2-2x) + (2a+2b)(-\frac{1}{2} + x)$$

Firstly, I have no idea how it arrived at this result. Second of all, I do not know how this would help prove linear independence for $\{f_1,f_2\}$. Any help would be appreciated.

Edit

Given that $\{f_1, f_2\}$ form the dual basis of a particular basis $\{v_1,v_2\}$ for some $v_1,v_2 \in V$, we know that $f_i(v_j) = \delta_{ij}$ where $\delta_{ij}$ is the Kronecker delta function. This implies that, letting $v_1 = p_1(x) = a_1 + b_1x$ and $v_2 = p_2(x) = a_2 + b_2x$ \begin{align*} 1 &= f_1(v_1) = a_1+\frac{b_1}{2} \\ 0 &= f_2(v_1) = 2a_1+2b_1 \end{align*}

Solving this system, we get that $v_1 = 2 - 2x$.

Similarly, \begin{align*} 0 &= f_1(v_2) = a_1+\frac{b_1}{2} \\ 1 &= f_2(v_2) = 2a_1+2b_1 \end{align*}

Solving gives us that $v_2 = -\frac{1}{2} + x$.

From this we get that the basis of $V$ for which $\{f_1, f_2\}$ is the dual basis is $\{(2 - 2x), (-\frac{1}{2} + x)\}$.

However, I am still struggling to prove linear independence in order to show that $\{f_1, f_2\}$ is a basis in the first place.

For $f_1,f_2$ to be linearly independent, you need to check that

$$c_1 f_1 + c_2 f_2 = 0_{V^{*}} \implies c_1 = c_2 = 0.$$

The element $c_1 f_1 + c_2 f_2$ is a linear functional so you need to check that if $c_1 f_1 + c_2 f_2$ is the zero functional then $c_1 = c_2 = 0$. This means that if

$$(c_1 f_1 + c_2 f_2)(v) = 0$$

for all polynomials $v = ax + b$ then you need to show that $c_1 = c_2 = 0$.

Let's write this explicitly. Assume that $c_1 f_1 + c_2 f_2 = 0_{V^{*}}$. Then for all $v = ax + b$ we have

$$c_1f_1(v) + c_2f_2(v) = c_1 \left( a + \frac{b}{2} \right) + c_2 \left( 2a + 2b \right) = 0.$$

Choosing $a = 1, b = -2$ we get

$$-4c_2 = 0 \implies c_2 = 0.$$

Similarly, choosing $a = 1, b = -1$, we get

$$\frac{1}{2} c_1 = 0 \implies c_1 = 0$$

so indeed the linear functionals $f_1,f_2$ are linearly independent.

The solution manual already gives you the dual basis. Recall that if $v_1,\dots,v_n$ is a basis for $V$ and $f^1, \dots, f^n$ is the corresponding dual basis of $V^{*}$, we have for any vector $v$ the identity

$$v = f^1(v) v_1 + \dots + f^n(v) v_n.$$

In your case, $v_1 = 2 - 2x$ and $v_2 = -\frac{1}{2} + x$.

• Isn't choosing a particular value for a and b getting rid of the generality of the equation. i.e. shouldn't linear independence hold for any values of a and b rather than some particular values of a and b. Also, did the solution manual arrive at these values for v1 and v2 through some particular method or were they only chosen due to their ability to form a vector v in V of the form a + bx when multiplied by the corresponding functional applied to v and put in linear combination? Mar 17 '18 at 19:40
• Ah, I think I may have answered my own comment here. You can plug in any values you want for a and b since the linear independence of functionals should hold for all v = a + bx. Therefore, they must hold for any arbitrarily chosen values for a and b. Choosing certain a's and b's allows us to get rid of c_1 or c_2 allowing us to solve for the other and showing that they are equal to 0. Mar 18 '18 at 0:17
• @OrrenRavid: They were chosen due to their ability to form a general vector in $V$ when multiplied by the corresponding functionals applied to $v$ and put in a linear combination, but this choice isn't arbitrary. Since $f_1,f_2$ are linearly independent, the vectors $v_1,v_2$ are determined uniquely by this condition. Mar 18 '18 at 0:30
• I agree that the choice isn't arbitrary when choosing $v_1$ and $v_2$ as I say in my edit to my original question. However, am I correct in my comment about the choices of a and b when proving linear independence? Mar 18 '18 at 0:58
• @OrrenRavid: Yep, you are correct. Mar 18 '18 at 2:12