I have this question that I took a shot at but I am not very familiar with Hermite or Laguerre, this my first time running across these type of polynomials and need some help please.
(a) The first four Hermite polynomials are $1, 2t,-2+4t^2,$ and $-12t+8t^3$. These polynomials arise naturally in the study of certain important differential equations in mathematical physics. Show that the first four Hermite polynomials form a basis of $\mathbb{P}_3$.
(b) The first four Laguerre polynomials are $1, 1-t, 2-4t + t^2,$ and $6-18t + 9t^2- t^3$. Show
that these four Laguerre polynomials form a basis of $\mathbb{P}_3$.
Results:
(a) The first four Hermite polynomials will be shown to form a basis of $\mathbb{P}_3$ by showing that they are linearly independent and that the number of polynomials equals the dimension of $\mathbb{P}_3$.
Consider the following linear combination of the four Hermite polynomials:
$x(1) + y(2t) + z(-2+4t^2) + w(-12t + 8t^3) = at^3 + bt^2 + ct + d$
The first four Hermite polynomials will be shown to be linearly independent by showing that the only linear combination of them that produces the zero polynomial ($0t^3+0t^2+0t+0$) is the trivial combination of zero times each polynomial.
That is all I could come about thus far with it. Can anyone refine or correct this if this is the wrong approach to the problem.
Best Answer
Forget all this Hermite and Laguerre stuff. The fact is that any family of polynomials with all different degrees is linearly free. Hence any family of polynomials with degrees $0$, $1$, $\ldots$, $n$ is a basis of the vector space of polynomials of degree at most $n$ (the space you denote by $\mathbb{P}_n$).
A relatively more sophisticated way of saying the same thing is that any triangular matrix with no zero on its diagonal is invertible.