$$ u_0, u_1, u_2 = 0, 1, 5$$
$$ p_0, p_1, p_2 = 1, 2, -6 $$
$$ \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 5 & 20 \\ \end{bmatrix} $$
I was supposed to find an interpolating polynomial using the collocation matrix above. The answer was this:
$$ 1 + u - 3/5(u^2 - u)$$ Now there was a question that would the polynomial be different if it were Lagrange polynomials? The answer suggested that no, they would not be different since in Lagrange gives us an identity matrix.
I am confused. If Lagrange gives us an identity matrix and I compute the polynomial in the same way that I did earlier, the answer would be different, no? It would be $$1 +2u -6(u^2 - u)$$ no?
Note that the Lagrange basis here consists of the following polynomials:
$$p_0(x)=\frac{(x-1)(x-5)}{(0-1)(0-5)}=\frac{(x-1)(x-5)}{5} \\ p_1(x)=\frac{x(x-5)}{(1)(1-5)}=\frac{x(x-5)}{-4} \\ p_2(x)=\frac{x(x-1)}{(5)(5-1)}=\frac{x(x-1)}{20}.$$
It is defined so that $p_i(x_i)=1$ and $p_i(x_j)=0$ for $j \neq i$. Then the interpolating polynomial in your case is $p_0(x)+2p_2(x)-6p_3(x)$. With this correction you do indeed get the same polynomial, though at first glance without simplifying it will not look the same as in the monomial basis or the Newton basis.