Exercise 6.2.2

The answers here might be different due to the different order of vectors chosen to be orthogonalized.

1.

Let

$$S=\{w_1=(1,0,1),w_2=(0,1,1),w_3=(1,3,3)\}.$$

Pick $v_1=w_1$. Then construct

$$v_2=w_2-\frac{⟨w_2,v_1⟩}{\parallel v_1{\text{∥}}^2}{v}_1$$

$$=w_2-\frac{1}{2}{v}_1=(-\frac{1}{2},1,\frac{1}{2}).$$

And then construct

$$v_3=w_3-\frac{⟨w_3,v_1⟩}{\parallel v_1{\text{∥}}^2}{v}_1-\frac{⟨w_3,v_2⟩}{\parallel v_2{\text{∥}}^2}{v}_2$$

$$=w_3-\frac{4}{2}{v}_1-\frac{4}{\frac{3}{2}}{v}_2=(\frac{1}{3},\frac{1}{3},-\frac{1}{3}).$$

As the demand in the exercise, we normalize $v_1$, $v_2$, and $v_3$ to be

$$u_1=(\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}),$$

$$u_2=(-\frac{1}{\sqrt{6}},\frac{4}{\sqrt{6}},\frac{1}{\sqrt{6}}),$$

$$u_3=(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}).$$

Let $\beta =\{u_1,u_2,u_3\}$. Now we have two ways to compute the Fourier coefficients of $x$ relative to $\beta$. One is to solve the system of equations

$$a_1{u}_1+a_2{u}_2+a_3{u}_3=x$$

and get

$$x=\frac{3}{\sqrt{2}}{u}_1+\frac{3}{\sqrt{6}}{u}_2+0u_3.$$

The other is to calculate the $i$-th Fourier coefficient

$$a_i=⟨x,u_i⟩$$

directly by Theorem 6.5. And the two consequences meet.

2.

Ur...don’t follow the original order. Pick $w_1=(0,0,1)$, $w_2=(0,1,1)$, and $w_3=(1,1,1)$ and get the answer

$$\beta =\{(0,0,1),(0,1,0),(1,0,0)\}$$

instantly. And easily we also know that the Fourier coefficients of $x$ relative to $\beta$ are $1,0,1$.

3.

The basis is

$$\beta =\{1,\sqrt{3}(2x-1),\sqrt{5}(6x^2-6x+1)\}.$$

And the Fourier coefficients are $\frac{3}{2},\frac{\sqrt{3}}{6},0$.

4.

The basis is

$$\beta =\{\frac{1}{\sqrt{2}}(1,i,0),\frac{1}{2\sqrt{17}}(1+i,1-i,4i)\}.$$

And the Fourier coefficients are $\frac{7+i}{\sqrt{2}},\sqrt{17}i$.

5.

The basis is

$$\beta =\{(\frac{2}{5},-\frac{1}{5},-\frac{2}{5},\frac{4}{5}),(-\frac{4}{\sqrt{30}},\frac{2}{\sqrt{30}},-\frac{3}{\sqrt{30}},\frac{1}{\sqrt{30}}),$$

$$(-\frac{3}{\sqrt{155}},\frac{4}{\sqrt{155}},\frac{9}{\sqrt{155}},\frac{7}{\sqrt{155}})\}.$$

And the Fourier coefficients are $10,3\sqrt{30},\sqrt{150}$.

6.

The basis is

$$\beta =\{(\frac{1}{\sqrt{15}},-\frac{2}{\sqrt{15}},-\frac{1}{\sqrt{15}},\frac{3}{\sqrt{15}}),(\frac{2}{\sqrt{10}},\frac{2}{\sqrt{10}},\frac{1}{\sqrt{10}},\frac{1}{\sqrt{10}}),$$

$$(-\frac{4}{\sqrt{30}},\frac{2}{\sqrt{30}},\frac{1}{\sqrt{30}},\frac{3}{\sqrt{30}})\}.$$

And the Fourier coefficients are $-\frac{3}{\sqrt{15}},\frac{4}{\sqrt{10}},\frac{12}{\sqrt{30}}$.

7.

The basis is

$$\beta =\{\left(\begin{array}{cc}\hfill \frac{1}{2}\hfill & \hfill \frac{5}{6}\hfill \\ \hfill -\frac{1}{6}\hfill & \hfill \frac{1}{6}\hfill \end{array}\right),\left(\begin{array}{cc}\hfill -\frac{\sqrt{2}}{3}\hfill & \hfill \frac{\sqrt{2}}{3}\hfill \\ \hfill \frac{1}{\sqrt{2}}\hfill & \hfill -\frac{1}{3\sqrt{2}}\hfill \end{array}\right),\left(\begin{array}{cc}\hfill \frac{1}{\sqrt{2}}\hfill & \hfill -\frac{1}{3\sqrt{2}}\hfill \\ \hfill \frac{\sqrt{2}}{3}\hfill & \hfill -\frac{\sqrt{2}}{3}\hfill \end{array}\right)\}.$$

And the Fourier coefficients are $24,6\sqrt{2},-9\sqrt{2}$.

8.

The basis is

$$\beta =\{\left(\begin{array}{cc}\hfill \frac{2}{\sqrt{13}}\hfill & \hfill \frac{2}{\sqrt{13}}\hfill \\ \hfill \frac{2}{\sqrt{13}}\hfill & \hfill \frac{1}{\sqrt{13}}\hfill \end{array}\right),\left(\begin{array}{cc}\hfill \frac{5}{7}\hfill & \hfill -\frac{2}{7}\hfill \\ \hfill -\frac{4}{7}\hfill & \hfill \frac{2}{7}\hfill \end{array}\right),\left(\begin{array}{cc}\hfill \frac{8}{\sqrt{373}}\hfill & \hfill -\frac{8}{\sqrt{373}}\hfill \\ \hfill \frac{7}{\sqrt{373}}\hfill & \hfill -\frac{14}{\sqrt{373}}\hfill \end{array}\right)\}.$$

And the Fourier coefficients are $5\sqrt{13},-14,\sqrt{373}$.

9.

The basis is

$$\beta =\{\frac{\sqrt{2}\sin <!--\; FUNCTION\; APPLICATION\; -->(t)}{\sqrt{\pi }},\frac{\sqrt{2}\cos <!--\; FUNCTION\; APPLICATION\; -->(t)}{\sqrt{\pi }},\frac{\pi -4\sin <!--\; FUNCTION\; APPLICATION\; -->(t)}{\sqrt{{\pi }^3-8\pi }},\frac{8\cos <!--\; FUNCTION\; APPLICATION\; -->(t)+2\mathit{\pi t}-{\pi }^2}{\sqrt{\frac{{\pi }^5}{3}-32\pi }}\}.$$

And the Fourier coefficients are $\frac{\sqrt{2}(2\pi +2)}{\sqrt{\pi }},-\frac{4\sqrt{2}}{\sqrt{\pi }},\frac{{\pi }^3+{\pi }^2-8\pi -8}{\sqrt{{\pi }^3-8\pi }},\frac{\frac{{\pi }^4-48}{3}-16}{\sqrt{\frac{{\pi }^5}{3}-32\pi }}$.

10.

The basis is

$$\{(\frac{1}{2^{\frac{3}{2}}},\frac{i}{2^{\frac{3}{2}}},\frac{2-i}{2^{\frac{3}{2}}},-\frac{1}{2^{\frac{3}{2}}}),(\frac{3i+1}{2\sqrt{5}},\frac{i}{\sqrt{5}},-\frac{1}{2\sqrt{5}},\frac{2i+1}{2\sqrt{5}}),$$

$$(\frac{i-7}{2\sqrt{35}},\frac{i+3}{\sqrt{35}},\frac{5}{2\sqrt{35}},\frac{5}{2\sqrt{35}})\}.$$

And the Fourier coefficients are $6\sqrt{2},4\sqrt{5},2\sqrt{35}$.

11.

The basis is

$$\{(-\frac{4}{\sqrt{47}},\frac{3-2i}{\sqrt{47}},\frac{i}{\sqrt{47}},\frac{1-4i}{\sqrt{47}}),(\frac{3-i}{2\sqrt{15}},-\frac{5i}{2\sqrt{15}},\frac{2i-1}{\sqrt{15}},\frac{i+2}{2\sqrt{15}}),$$

$$(\frac{-i-17}{2\sqrt{290}},\frac{8i-9}{2\sqrt{290}},\frac{8i-9}{\sqrt{290}},\frac{8i-9}{2\sqrt{290}})\}.$$

And the Fourier coefficients are $-\sqrt{47}i-\sqrt{47},4\sqrt{15}i-2\sqrt{15},2\sqrt{290}i+2\sqrt{290}$.

12.

The basis is

$$\{\left(\begin{array}{cc}\hfill \frac{1-i}{2\sqrt{10}}\hfill & \hfill \frac{-3i-2}{2\sqrt{10}}\hfill \\ \hfill \frac{2i+2}{2\sqrt{10}}\hfill & \hfill \frac{i+4}{2\sqrt{10}}\hfill \end{array}\right),\left(\begin{array}{cc}\hfill \frac{3\sqrt{2}i}{5}\hfill & \hfill \frac{-i-1}{5\sqrt{2}}\hfill \\ \hfill \frac{1-3i}{5\sqrt{2}}\hfill & \hfill \frac{i+1}{5\sqrt{2}}\hfill \end{array}\right),\left(\begin{array}{cc}\hfill \frac{-43i-2}{5\sqrt{323}}\hfill & \hfill \frac{1-21i}{5\sqrt{323}}\hfill \\ \hfill -\frac{68i}{5\sqrt{323}}\hfill & \hfill \frac{34i}{5\sqrt{323}}\hfill \end{array}\right)\}.$$

And the Fourier coefficients are $2\sqrt{10}-6\sqrt{10}i,10\sqrt{2},0$.

13.

The basis is

$$\{\left(\begin{array}{cc}\hfill \frac{i-1}{3\sqrt{2}}\hfill & \hfill -\frac{i}{3\sqrt{2}}\hfill \\ \hfill \frac{2-i}{3\sqrt{2}}\hfill & \hfill \frac{3i+1}{3\sqrt{2}}\hfill \end{array}\right),\left(\begin{array}{cc}\hfill -\frac{4i}{\sqrt{246}}\hfill & \hfill \frac{-9i-11}{\sqrt{246}}\hfill \\ \hfill \frac{5i+1}{\sqrt{246}}\hfill & \hfill \frac{1-i}{\sqrt{246}}\hfill \end{array}\right),\left(\begin{array}{cc}\hfill \frac{-118i-5}{\sqrt{39063}}\hfill & \hfill \frac{-26i-7}{\sqrt{39063}}\hfill \\ \hfill -\frac{145i}{\sqrt{39063}}\hfill & \hfill -\frac{58}{\sqrt{39063}}\hfill \end{array}\right)\}.$$

And the Fourier coefficients are $3\sqrt{2}i+6\sqrt{2},-\sqrt{246}i-\sqrt{246},0$.