Exercise:

Let \(V\) be the space of real polynomial functions of degree at most \(3\). Consider the quadratic form \(q_1:V \to \mathbb{R}\) given by $$ q_1(f):=|f(-1)|^2+|f(0)|^2+|f(1)|^2. $$ Is this form positive definite?

Consider another form \(q:V\to \mathbb{R}\) given by $$ q(f):=\int_{0}^\infty e^{-x} |f(x)|^2 dx $$ Diagonalize the form using Gram-Schmidt procedure starting with the standard monomial basis \(\{1,x,x^2,x^3\}\).

Solution:

In order for \(q_1\) be positive definite, it should vanish only for the zero polynomial, namly \(q_1(f)=0\) if and only if \(f(x)=0\). However, the polynomial \(g(x) = x(x-1)(x+1) \in V\) is nonzero and yet has $$ q_1(g) = |g(-1)|^2+|g(0)|^2+|g(1)|2 = 0^2+0^2+0^2 = 0,$$ hence \(q_1\) is not positive definite on \(V\).

For the second part, note the bi-linear form corresponding to the quadratic form: $$ Q(f,g) = \int_{0}^\infty e^{-x} f(x)g(x) dx $$ and we will utilize the identity \( \int_0^\infty e^{-x} x^n = n! \) for any integer \(n\geq 1\).

Starting with the standard basis is \(f_k(x) = x^k\), the first Gram-Schmidt vector is simply be $$ \begin{align} e_0(x) & = f_0(x) = \boxed{1} \\ \Rightarrow ~~Q(e_0,f_k) &= \int_0^\infty e^{-1} x^k dx = k! \\ Q(e_0,e_0) &= \int_0^\infty e^{-1} 1 dx = 1 \end{align} $$ The second Gram-Schmidt vector follows from that: $$ \begin{align} e_1(x) &= f_1(x) – \frac{Q(e_0,f_1)}{Q(e_0,e_0)}e_0(x) = \boxed{x – 1}, \\ \Rightarrow ~~Q(e_1,e_1) &= \int_0^\infty e^{-1} (x-1)^2 dx = \int_0^\infty e^{-x}(x^2 – 2x +1) dx = 2!-2+1 = 1\\ Q(e_1,f_k) &= \int_0^\infty e^{-1} (x-1)x^k dx = (k+1)!-k! = k\cdot k! \end{align} $$ The third vector: $$ \begin{align} e_2(x) &= f_2(x) – \frac{Q(e_1,f_2)}{Q(e_1,e_1)}e_1(x)- \frac{Q(e_0,f_2)}{Q(e_0,e_0)}e_0(x) = x^2 – 4 (x-1) -2 = \boxed{x^2 -4x + 2}. \\ \Rightarrow ~~Q(e_2,e_2) &= \int_0^\infty e^{-1} (x^2-4x+2)^2 dx = \int_0^\infty e^{-1}(x^4-8x^3 +20x^2 -16x + 4)dx = 4 \\ Q(e_2,f_k) &= \int_0^\infty e^{-1} (x^2-4x+2)x^k dx = (k+2)!-4(k+1)! +2k! \end{align} $$ And the last is: $$ \begin{align} e_3(x) &= f_3(x) – \frac{Q(f_3,e_2)}{Q(e_2,e_2)}e_2(x)- \frac{Q(e_1,f_3)}{Q(e_1,e_1)}e_1(x)- \frac{Q(e_0,f_3)} {Q(e_0,e_0)}e_0(x) \\ &= x^3 – \frac{36}{4}(x^2-4x+2) – \frac{3\cdot 3!}{1}(x-1) – 3! = \boxed{x^3-9x^2 +18x -6} \end{align} $$

The diagonalized basis \(\{e_0=1,~e_1=x-1,~e_2= x^2-4x+2,~e_3=x^3-9x^2 +18x -6\}\) is part of the renowned Laguerre polynomials, which are (among other things) orthogonal with respect to the weighted inner product defined by \(Q(f,g)\) above. That is, for every polynomial \(p(x) = \sum_{k=0}^3 a_k e_k(x) \) we have \( q(p) = \sum_{k=0}^3 \frac{1}{q(e_k)} a_k^2 \)

Exercise: Find the signature of the quadratic form \(q_1(f) =|f(-1)|^2+|f(0)|^2+|f(1)|^2.\) above.

Solution:

First note that being a sum of non-negative numbers, the form \(q_1(f)\geq 0\) (i.e no vectors have negative form value) is non-negative definite (sometimes called positive semidefinite). Its signature has \(n_{-} = 0 \) Our next step is to find \(n_0\) – the dimension of the nullspace of the quadratic form \(q_1\). If \(q_1(f)=0\) then it must have three roots at \ (x=\pm 1,0\), hence $$ f(x) = h(x) x(x-1)(x+1). $$ However, as we require that \(f\in V\), or \(\text{deg}(f) \leq 3 \), the factor \(h(x)\) must be some real constant \(h(x) = \alpha\), hence all nullvectors of \(q_1(f)=0\) have the form $$ f(x) = \alpha x(x-1)(x+1), $$ forming a one dimensional span, hence \(n_0 = 1\). Lastly, we compute \(n_{+}\): since \( \text{dim}(V) = 4 = n_{+} + n_0 + n_{-} \) we have \(n_{+} = 3\).

Note: the form \(q_1\) is positive definite on the space of polynomials of degree \(\leq 2\)

# solutions to 2.14

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