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Let $V$ be the vector space of all polynomial functions from $\mathbb{R}$ into $\mathbb{R}$ which have degree less than or equal to $2.$ Let $t_1,t_2$ and $t_3$ be any three distinct real numbers. Then show that $\{p_1,p_2,p_3\}$ forms a basis for $V$ where \begin{align} p_1(x)=\frac{(x-t_2)(x-t_3)}{(t_1-t_2)(t_1-t_3)}\\ p_2(x)=\frac{(x-t_1)(x-t_3)}{(t_2-t_1)(t_2-t_3)}\\ p_3(x)=\frac{(x-t_1)(x-t_2)}{(t_3-t_1)(t_3-t_2)} \end{align}

Showing these vectors are linearly independent is trivial. However showing that any given polynomial $c_1+c_2x+c_3x^2$ is in the span amounts to solving

$$\begin{pmatrix}\frac{t_2t_3}{(t_1-t_2)(t_1-t_3)}&\frac{t_1t_3}{(t_2-t_1)(t_2-t_3)}&\frac{t_1t_2}{(t_3-t_1)(t_3-t_2)}\\\\\frac{-t_2-t_3}{(t_1-t_2)(t_1-t_3)}&\frac{-t_1-t_3}{(t_2-t_1)(t_2-t_3)}&\frac{-t_1-t_2}{(t_3-t_1)(t_3-t_2)}\\\frac{1}{(t_1-t_2)(t_1-t_3)}&\frac{1}{(t_2-t_1)(t_2-t_3)}&\frac{1}{(t_3-t_1)(t_3-t_2)}\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}c_1\\c_2\\c_3\end{pmatrix}.$$

It seems that the coefficient matrix is invertible. But this is so humdrum.

Is there any better shorter way to do it?

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$\{1, x, x^2\}$ is a basis for $V$, so your vector space is 3-dimensional. Then it is enough to show that 3 vectors are linearly independent.

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  • $\begingroup$ Argh! Why on earth did I not realize this earlier. Thanks $\endgroup$ – Bijesh K.S Jun 29 '17 at 12:58
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Just for the sake of doing it: It is also easy to show that $p_1,p_2,p_3$ are a generating set without using the dimension argument:

If $f$ is any polynomial of degree at most two, then $f$ and $f(t_1)p_1+f(t_2)p_2+f(t_3)p_3$ agree at the three distinct values $t_1, t_2, t_3$. Since they are both of degree at most two, they are equal. This shows that any such $f$ is a linear combination of $p_1,p_2,p_3$.

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$ \dim V=3$ and $\{p_1,p_2,p_3\}$ is a basis of $V$ imply that $span (\{p_1,p_2,p_3\})=V.$

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