User:J D Bowen/Math725 HW8

1) For $c_i\in\mathbb{C} \ $, define $\langle \vec{v},\vec{w} \rangle_w =\Sigma_i c_i v_i \overline{w_i} \ $.

Observe that if $\langle *,* \rangle_w \ $ is an inner product $\iff \ $ it is real and positive for any two vectors, linear in the first term, and the zero vector taken with itself has product 0. Since $\langle \vec{v},\vec{w} \rangle_w=c_i \langle \vec{v},\vec{w} \rangle \ $, the latter conditions is always true, but the first holds if and only if $c_i>0 \ $, since $c_i>0 \iff c_i * R \in\mathbb{R}_+ \ $, where $R \ $ is any nonnegative real (from the regular inner product).

2) Let a matrix $M\in \mathfrak{L}(\mathbb{R}^n,\mathbb{R}^n) \ $ have rows $\vec{r}_1,\dots,\vec{r}_n \ $. Observe that $M\vec{v}=\vec{0} \iff \langle\vec{v},\vec{r}_j\rangle = 0 \forall j \iff \vec{v}\perp \vec{r}_j \forall j \iff \vec{v}\in\text{span}(\vec{v}_1, \dots, \vec{v}_n) \ $.

3) Define $\langle f,g \rangle = \int_{-1}^1 f(t)g(t)dt \ $. Observe that

$\langle f+g, h \rangle = \int_{-1}^1 (f(t)+g(t))h(t)dt =\int_{-1}^1 (f(t)h(t)+g(t)h(t) )dt =\int_{-1}^1 f(t)h(t)dt+\int_{-1}^1g(t)h(t)dt = \langle f, h \rangle+\langle g,h \rangle \ $,

$\langle af, g \rangle = \int_{-1}^1 af(t)g(t)dt = a\int_{-1}^1 f(t)g(t)dt = a\langle f,g\rangle \ $

$\langle f,g \rangle = 0 \iff \int_{-1}^1 f(t)g(t)dt=0 \iff f\lor g = 0 \ $,

with the last double implication true because the functions are polynomials.

So this defines an inner product. We have

$||x||^2 = \int_{-1}^1 t^2dt = \frac{1^3}{3}-\frac{(-1)^3}{3}= \frac{2}{3} \implies ||x||=\sqrt{\frac{2}{3}} \ $,

$||x^n||^2 = \int_{-1}^1 t^{2n}dt = \frac{2}{2n+1} \implies ||x^n||=\sqrt{\frac{2}{2n+1}} \ $

We aim for a constant function, a linear function, and a quadratic function $(c,p,q) \ $ which are orthonormal. Begin with $\left\{{1,x,x^2}\right\} \ $ and apply the Gram-Schmidt process. Define $c(x)=1 \ $.

Then we define

$p(x)=\frac{x-\frac{\langle x,1 \rangle}{\langle 1,1 \rangle }}{||x-\frac{\langle x,1 \rangle}{\langle 1,1 \rangle }||} = x\sqrt{\frac{3}{2}} \ $.

Finally, define

$q(x)=\frac{x^2-\frac{\langle x^2,1\rangle}{\langle 1,1\rangle}1-\frac{\langle x^2,x\sqrt{\frac{3}{2}}\rangle}{\langle x\sqrt{\frac{3}{2}},x\sqrt{\frac{3}{2}}\rangle}x\sqrt{\frac{3}{2}}}{||x^2-\frac{\langle x^2,1\rangle}{\langle 1,1\rangle}1-\frac{\langle x^2,x\sqrt{\frac{3}{2}}\rangle}{\langle x\sqrt{\frac{3}{2}},x\sqrt{\frac{3}{2}}\rangle}x\sqrt{\frac{3}{2}}||}

= \frac{ x^2-\frac{1}{\sqrt{6}}

}{||x^2-\frac{1}{\sqrt{6}}||} = \frac{x^2-\frac{1}{\sqrt{6}}}{ \sqrt{2\left({ \frac{1}{5}-\frac{2}{3\sqrt{6}}+\frac{1}{6} }\right) }}

= (x^2-\frac{1}{\sqrt{6}})\sqrt{ \frac{90}{51-20\sqrt{6}}} \ $

We are guaranteed these are orthonormal by the construction.

4) Observe that

$||\vec{v}+\vec{u}||^2 = ||\vec{v}||^2+||\vec{u}||^2 \iff \langle \vec{v}+\vec{u},\vec{v}+\vec{u} \rangle = \langle \vec{v},\vec{v} \rangle + \langle \vec{u},\vec{u} \rangle \ $

5) Define $f(t)=\langle t\vec{u},\vec{v} \rangle \ $. Since these two vectors are fixed, $\langle \vec{u},\vec{v} \rangle \ $ is a constant.  Since $\langle t\vec{u},\vec{v} \rangle=t\langle \vec{u},\vec{v} \rangle \ $, we have $f'(t)=\langle \vec{u},\vec{v} \rangle \ $

Observe that $\langle \vec{v},\vec{w}t \rangle = \overline{t}\langle \vec{v},\vec{w} \rangle \ $. Since $t\mapsto \overline{t}k \ $, and since the complex conjugate is not holomorphic, $t\mapsto \langle \vec{v},\vec{w}t \rangle \ $ has no complex derivative.