Given a function $f$, and information on $n$ different points: $\{f(x_i),f'(x_i):i=1,\ldots,n\}$, we can get an interpolant polynomial of degree $2n-1$ by Hermite interpolant. This is equivalent to solve a linear equation: $$ \pmatrix{1&t&t^2&\cdots &t^{2n-1}\\0&1&2t&\cdots&(2n-1)t^{2n-2}}\textbf{c}=\pmatrix{f(t)\\f'(t)}$$ where $t=(x_1,\ldots,x_n)^T$
It can be expected that the matrix on the left, say $T\in \mathbb{R}^{2n\times 2n}$, is nonsingular, so that we can always find the solution $\textbf{c}$ and obtain the interpolant of degree $2n-1$.
My question is: how to prove $T$ is nonsingular? I think the only thing I can do is to try to prove the linear independence of the column vectors.
I get \begin{cases} a_0+a_1 x_1+a_2 x_1^2+\cdots+a_{2n-1}x_1^{2n-1}=0\\ \qquad\qquad\qquad\vdots\\ a_0+a_1 x_n+a_2 x_n^2+\cdots+a_{2n-1}x_n^{2n-1}=0\\ 0+a_1+2 a_2 x_1+\cdots+(2n-1)x_1^{2n-2}=0\\ \qquad\qquad\qquad\vdots\\ 0+a_1+2 a_2 x_n+\cdots+(2n-1)x_n^{2n-2}=0 \end{cases} I have no idea what to do next.
I know the existence of Hermite interpolation is of course true, but I wonder if we can prove it through this linear algebra problem.