Answer
see work step by step
Work Step by Step
(a) By supposition, we know that
\[
\left[\begin{array}{ccccc}
1 & x_{1} & x_{1}^{2} & \cdots & x_{1}^{n-1} \\
1 & x_{2} & x_{2}^{2} & \cdots & x_{2}^{n-1} \\
\vdots & \vdots & \vdots & & \vdots \\
1 & x_{n} & x_{n}^{2} & \cdots & x_{n}^{n-1}
\end{array}\right]\left[\begin{array}{c}
c_{0} \\
c_{1} \\
\vdots \\
c_{n-1}
\end{array}\right]=\left[\begin{array}{c}
y_{1} \\
y_{2} \\
\vdots \\
y_{n}
\end{array}\right]
\]
Let's multiply out the matrices on the left-hand side of this equation to get
\[
\left[\begin{array}{c}
c_{0}+c_{1} x_{1}+c_{2} x_{1}^{2}+\cdots+c_{n-1} x_{1}^{n-1} \\
c_{0}+c_{1} x_{2}+c_{2} x_{2}^{2}+\cdots+c_{n-1} x_{2}^{n-1} \\
\vdots \\
c_{0}+c_{1} x_{n}+c_{2} x_{n}^{2}+\cdots+c_{n-1} x_{n}^{n-1}
\end{array}\right]=\left[\begin{array}{c}
y_{1} \\
y_{2} \\
\vdots \\
y_{n}
\end{array}\right]
\]
Comparing the rows on the left-hand side of this equation to $p(t),$ we can see that
\[
\left[\begin{array}{c}
p\left(x_{1}\right) \\
p\left(x_{2}\right) \\
\vdots \\
p\left(x_{n}\right)
\end{array}\right]=\left[\begin{array}{c}
y_{1} \\
y_{2} \\
\vdots \\
y_{n}
\end{array}\right]
\]
From which we conclude that $p\left(x_{i}\right)=y_{i}$ for each $i=1,2, \ldots, n$
We note that the Fundamental Theorem of Algebra FTA tells us that the maximum number of real zeros that a degree $n$ polynomial can have is $n$
Notice that any nonzero solution $\mathbf{c}$ to the homogeneous equation $V \mathbf{c}=$ 0 will be the coefficients of some degree $n-1$ polynomial given by
\[
p(t)=c_{0}+c_{1} t+\cdots+c_{n-1} t^{n-1}
\]
But then, from part (a), we know that all of the $n$ numbers $x_{1}, \ldots, x_{n}$ are zeros of this polynomial*.
From the FTA we know that it's not possible for a degree $n-1$ or small polynomial to have $n$ zeros. Hence the only solution to this equation is the trivial solution $(\mathbf{c}=\mathbf{0}) .$ The Invertible Matrix Theorem (IMT) then tells us that the columns of $V$ are linearly independent.
$*$ Set all of the $y$ 's in part (a) to 0 to see that $p\left(x_{i}\right)=0$ for all $i .$ I.e. each $x_{i}$ is a zero of the polynomial $p$
(c) From part (b), we know that if $x_{1}, \ldots, x_{n}$ are distinct numbers, then the columns of $V$ are linearly independent and from the IMT we then see that $V$ is invertible. since $V$ is invertible, there is a solution $\mathbf{c}$ to the equation $V \mathbf{c}=\mathbf{y}$ from every $\mathbf{y}$ in $\mathbb{R}^{n}$
From part (a) we know that a solution to $V \mathbf{c}=\mathbf{y}$ are the coefficients of an interpolating polynomial of degree $n-1$ (or smaller) for the points $\left(x_{1}, y_{1}\right), \ldots,\left(x_{n}, y_{n}\right)$
From part (a) we know that a solution to $V \mathbf{c}=\mathbf{y}$ are the coefficients of an interpolating polynomial of degree $n-1$ (or smaller) for the points $\left(x_{1}, y_{1}\right), \ldots,\left(x_{n}, y_{n}\right)$
Hence we find that if $x_{1}, \ldots, x_{n}$ are distinct numbers, then for any numbers $y_{1}, \ldots, y_{n},$ there is an interpolating polynomial of degree $n-1$ or smaller, as required.
