$V = \operatorname{span}(S)$, where $S = \{(1, i, 0), (1 – i, 2, 4i)\}$, and $x = (3 + i, 4i, -4)$.
- Apply the Gram–Schmidt process to the given subset $S$ of the inner
product space $V$ to obtain an orthogonal basis for $\operatorname{span}(S)$.- Normalize the vectors in this basis to obtain an orthonormal basis $B$ for $\operatorname{span}(S)$.
- Compute the Fourier coefficients of the given vector relative to $0$ by solving the system of equations $a_1u_1 + a_2u_2 + a_3u_3 = x$, where the $u_j$ are the vectors of basis $B$.
- Compute the Fourier coefficients directly by $a_i = \langle x, u_i\rangle$.
I'm pretty sure I know how to, and I did the first two parts correctly; the computed orthonormal basis is:
$$B = \left\{ \frac{1}{\sqrt2}(1, i, 0), \frac{1}{\sqrt{17}}(1 + i, 1 – i, 4i)\right\}.$$
The correct answer for the Fourier coefficients is
$$\frac{7 + i}{\sqrt 2}, \sqrt{17} i.$$
But I can't seem to get either of them through either method. I can't get the answer through (3), but I feel it's just a calculation error. I am really not sure how to compute (4) though — can someone show me the step-by-step computation to get the Fourier coefficient through (4)?
Also, if my orthonormal basis is incorrect, please show me your answer too.
Best Answer
In follow,
with the star symbol, I mean complex conjugate, i.e. for example $(3-2i,3,2i)^*=(3+2i,3,-2i)$,
If $v=\{{v_1,v_2,v_3}\}$ and $u=\{{u_1,u_2,u_3}\}$ , we define
$v.u=v_1u_1+v_2 u_2+v_3 u_3$ and $<v,u>=u^*.v$.
The detailed answer is as follow,
1.
$v_1=(1,i,0)$ and $v_2=(1-i,2,4i)$, using the Gram Schmidt process, we have
$$w_1=v_1=(1,i,0)$$ $$w_2=v_2-\frac{v_1^*.v_2}{v_1^*.v_1}v_1=(1-i,2,4i)-\frac{(1,-i,0).(1-i,2,4i)}{(1,-i,0).(1,i,0)}(1,i,0)\\ =(1-i,2,4i)-\frac{1-3i}{2}(1,i,0)=(\frac{1+i}{2},\frac{1-i}{2},4i)$$
2. $$u_1=\frac{{w}_1}{\sqrt{{w}_1^*.{w}_1}}=\frac{(1,i,0)}{\sqrt{(1,-i,0).(1,i,0)}}=\frac{1}{\sqrt{2}}(1,i,0)$$ $$u_2=\frac{{w}_2}{\sqrt{{w}_2^*.{w}_2}}=\frac{(\frac{1+i}{2},\frac{1-i}{2},4i)}{\sqrt{(\frac{1-i}{2},\frac{1+i}{2},-4i).(\frac{1+i}{2},\frac{1-i}{2},4i)}}\\=\frac{1}{\sqrt{17}}(\frac{1+i}{2},\frac{1-i}{2},4i)=\frac{1}{2\sqrt{17}}(1+i,1-i,8i)$$
3.
Since $x \in {C^3}$, we need three orthonormal basis $\{{u_1,u_2,u_3}\}$. The unit vector $u_3$ must be orthogonal to both $u_1$ and $u_2$, therefore, if we assume $u_3=(\alpha_1,\alpha_2,\alpha_3)$ then,
$u_1^*.u_3=0 \to \frac{1}{\sqrt{2}}(1,-i,0).(\alpha_1,\alpha_2,\alpha_3)=0 \to \alpha_1-i\alpha_2=0 \,\,(1)$ $u_2^*.u_3=0 \to \frac{1}{2\sqrt{17}}(1-i,1+i,-8i).(\alpha_1,\alpha_2,\alpha_3)=0 \to (1-i)\alpha_1+(1+i)\alpha_2-8i\alpha_3=0 \,\,(2)$
Equations $(1)$ and $(2)$ have solutions of the form of $\alpha (4i,4,1 - i)$. Using the normalization condition, $$\sqrt {{{\left| {{\alpha _1}} \right|}^2} + {{\left| {{\alpha _2}} \right|}^2} + {{\left| {{\alpha _3}} \right|}^2}} = 1$$ we obtain, $$u_3=\frac{1}{\sqrt{34}}(4i,4,1-i)$$
Now, we turn into calculation of Fourier coefficients. From vector equation $$a_1u_1+a_2u_2+a_3u_3=x$$ or $$\frac{a_1}{\sqrt{2}}(1,i,0)+\frac{a_2}{2\sqrt{17}}(1+i,1-i,8i)+\frac{a_3}{\sqrt{34}}(4i,4,1-i)=(3+i,4i,-4)$$ we have these linear system of equations, $$\frac{a_1}{\sqrt{2}}+\frac{a_2}{2\sqrt{17}}(1+i)+\frac{4ia_3}{\sqrt{34}}=3+i$$ $$\frac{i a_1}{\sqrt{2}}+\frac{a_2}{2\sqrt{17}}(1-i)+\frac{4a_3}{\sqrt{34}}=4i$$ $$\frac{8ia_2}{2\sqrt{17}}+\frac{a_3}{\sqrt{34}}(1-i)=-4$$ which can be solved easily, $$a_1=\frac{7+i}{\sqrt{2}}$$ $$a_2=\sqrt{17}i$$ $$a_3=0$$
4. Using given equation, we have $$a_1=u_1^*.x=\frac{1}{\sqrt{2}}(1,-i,0).(3+i,4i,-4)=\frac{7+i}{\sqrt{2}}$$ $$a_2=u_2^*.x=\frac{1}{2\sqrt{17}}(1-i,1+i,-8i).(3+i,4i,-4)=\sqrt{17}i$$ $$a_3=u_3^*.x=\frac{1}{\sqrt{34}}(-4i,4,1+i).(3+i,4i,-4)=0$$