I'd appreciate some help with this problem! My professor in my graduate linear algebra class is giving us some tough homework.
$$P(x) = p(x) = a_0 + a_1x + .... + a_nx^{n}$$ such that $a_i$ is real numbers
a.) Show that $P(x)$ is a vector space under the pointwise definition of vector addition and scalar multiplication.
b.) Show that the set $Q = \{q(x) = x^k$ such that $k = 0,1,....,n\}$ of monomials of order $k \le n$ form a basis for $P(x)$.
c.) Fix a set of a distinct points $x_0 < x_1 < ... < x_n$ in the reals and define the mapping
$K: p(x)| -> (p(x_0) p(x_1) ... p(x_n))$ Note: (this is a column vector) From P(x) to the reals of dimension n + 1. Show that this mapping has no kernal in P(x) and must be invertible.
d.) Using the formula for Lagrange polynomials of order n, calculate the column vector $K_{c_k}$ for each $k = 0,\dots,n$, and use this result to give an inverse mapping $K^{-1}$. Reals of dimension $n -> P(x)$.
There's just a lot here and it's getting to me. A little bit of guidance would be great!
I present hints and then solutions.
Hints: a. You need to show adding two polynomials of order $<=n$ yields a polynomial of order $<=n$. Also you need to show multiplying a polynomial of order $n$ with any real number yields a polynomial of order $<=n$
b. You need to show that all polynomials of order $<=n$ is a linear combination of monomials in $Q$.
c. Kernel of the map are elements that would be mapped to zero. In order for that to happen, we would be dealing with roots of $p(x)$. How many roots should $p(x)$ have?
d. I don't know what $c_k$ is but recall that a Lagrange polynomial is a polynomial that definitely goes through a prescribed number of points. Those prescribed number of points would be the column vector in $\mathbb{R}^{n+1}$ space.
Solutions: a. $\sum_{i=0}^n a_i x^i + \sum_{i=0}^n b_i x^i = \sum_{i=0}^n (a_i+b_i)x^i$, $\alpha \sum_{i=0}^n a_i x^i = \sum_{i=0}^n (\alpha a_i) x^i$.
b. straight forward. pretty much by definition.
c. $p(x)$ has at most $n$ distinct roots by the Fundamental Theorem of Algebra, therefore one of the $n+1$ numbers $p(x_0), p(x_1), ... $ must be non-zero, hence the map has trivial kernel. (it cannot have no kernel, because the $0$ polynomial is in the kernel )
d. Just plug in the $n+1$ points into the Lagrange polynomial formula, I'm sure you saw it in class.