Understanding example 22 of duality from Hoffman and Kunze's Linear Algebra

Question

I encountered one example in Hoffman and Kunze's Linear Algebra.
I did not get why one step happens, so I couldn't proceed further.

I do not get from where the highlighted portion came.
Any help will be appreciated.

Answer 1

$L = 0$ means that for any polynomial $f \in V$ we have: $$L(f) = c_1L_1(f)+L_2(f)+c_3L_3(f) = c_1f(t_1)+c_2f(t_2)+c_3f(t_3)=0$$ In particular setting $f(x) = 1$ gives $c_1+c_2+c_3 =0$, $f(x)=x$ gives $c_1t_1+c_2t_2+c_3t_3 =0$ and $f(x) = x^2$ gives $c_1t_1^2+c_2t_2^2+c_3t_3^2 =0$.

Answer 2

We want to check the following: given the real vector space $V = \{ p : \mathbb{R} \to \mathbb{R} : p \text{ is a polynomial function} \}$ and real numbers $t_1$, $t_2$ and $t_3$, the linear functionals $L_i : V \to \mathbb{R}$ defined by $$ L_i(p) = p(t_i), \qquad i = 1,2,3, $$ are linearly independent. So, we want to check that if the linear functional $L$ given by $$ L = c_1 L_1 + c_2 L_2 + c_3 L_3 $$ is the zero functional, then each $c_i$ must be zero. Now, $L = 0$ means that $L(p) = 0$ for any $p \in V$. In particular, $L(p_i)$ must be zero for each of the polynomial functions $p_1(x) = 1$, $p_2(x) = x$ and $p_3(x) = x^2$. Evaluating $L$ at each of these polynomial functions, we get the highlighted equations: \begin{align} L(p_1) &= c_1L_1(p_1) + c_2 L_2(p_1) + c_3L_3(p_1)\\ &= c_1 p_1(t_1) + c_2 p_1(t_2) + c_3 p_1(t_3) \\ &= c_1 + c_2 + c_3.\\ & \\ L(p_2) &= c_1L_1(p_2) + c_2 L_2(p_2) + c_3L_3(p_2)\\ &= c_1 p_2(t_1) + c_2 p_2(t_2) + c_3 p_2(t_3) \\ &= c_1 t_1 + c_2 t_2 + c_3 t_3.\\ & \\ L(p_3) &= c_1L_1(p_3) + c_2 L_2(p_3) + c_3L_3(p_3)\\ &= c_1 p_3(t_1) + c_2 p_3(t_2) + c_3 p_3(t_3) \\ &= c_1 t_1^2 + c_2 t_2^2 + c_3 t_3^2.\\ \end{align} So, we have the highlighted equations, \begin{alignat}{10} c_1 & {}+{} & c_2 & {}+{} & c_3 & {}={} 0 \\ t_1 c_1 & {}+{} & t_2 c_2 & {}+{} & t_3 c_3 & {}={} 0 \\ t_1^2 c_1 & {}+{}& t_2^2 c_2 & {}+{} & t_3^2 c_3 & {}={} 0. \end{alignat}