# Linear independece of linear functionals on vector space of all polynomials

This is from an example of a textbook that I didn't totally understand.

Let's suppose $V$ is the vector space of all polynomials functions from $\mathbb R$ into $\mathbb R$ which have degree less than or equal to 2.
Let $t_1$, $t_2$ and $t_3$ three distinct real numbers and let $$L_i(p) = p(t_i)$$

Then $L_1$, $L_2$ and $L_3$ are linear functionals on $V$ and they are linear independent.

Now, for proving that statement I put $$0 = c_1L_1+c_2L_2+c_3L_3$$ for all $c_1$, $c_2$ and $c_3$ and for each $p$ in $V$, and this should give us $c_1 = c_2 = c_3 = 0$
How can I prove that for all $p$ in $V$? I was thinking on considering a basis for the polynomials but then I'll introduce new scalars different from $c_1$, $c_2$ and $c_3$ and I got things messed up. Thanks in advance.

You need to show that

$$0=c_1p(t_1)+c_2p(t_2)+c_3p(t_3),\quad \forall p\in V$$

only holds when $c_1=c_2=c_3=0$. By example: define polynomials $p_k$ such that $p_k(t_j)=1-\delta_{k,j}$ where $\delta_{k,j}$ is the Kronecker delta (these polynomials can be defined explicitly using the Lagrange polynomials, if you need to prove it existence).

Then you have an homogeneous system of equations

$$0=c_1+c_2,\quad 0=c_1+c_3,\quad 0=c_2+c_3$$

what imply that $-c_1=c_2=c_3$, so all constant are zero.$\Box$

• Sorry, if I define particular polynomials like those, the assertion $\forall p\in V$ still holds? – mate89 Jun 23 '17 at 4:21
• the unique solution $c_1=c_2=c_3=0$ holds for any polynomial $p$ – Masacroso Jun 23 '17 at 13:57