Orthonormal Basis

[nasi112]

I don't understand why

[MATH]\left(v_1,\frac{v_1 + v_2}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}}[(v_1, v_1) + (v_1, v_2)][/MATH]
when I simplify the right hand side and compare it with the left side, I get

[MATH]v_1 = \frac{v_1 + v_1}{\sqrt{2}} = \frac{2v_1}{\sqrt{2}} = \sqrt{2}v_1[/MATH]

 

[lex]

The notation [MATH](\boldsymbol{a},\boldsymbol{b})[/MATH] represents the 'dot product', i.e. [MATH]\boldsymbol{a}.\boldsymbol{b}[/MATH]Now [MATH]\boldsymbol{v_1}.\boldsymbol{v_1}=\binom{1}{0}.\binom{1}{0}=1[/MATH]and [MATH]\boldsymbol{v_1}.\boldsymbol{v_2}=\binom{1}{0}.\binom{0}{1}=0[/MATH]so e.g. [MATH]\left(\boldsymbol{v_1},\frac{\boldsymbol{v_1}+\boldsymbol{v_2}}{\sqrt{2}} \right) = \boldsymbol{v_1}. \frac{1}{\sqrt{2}}(\boldsymbol{v_1}+\boldsymbol{v_2})=\frac{1}{\sqrt{2}}\boldsymbol{v_1}.(\boldsymbol{v_1}+\boldsymbol{v_2})= \frac{1}{\sqrt{2}}(\boldsymbol{v_1}.\boldsymbol{v_1}+\boldsymbol{v_1}.\boldsymbol{v_2})=\frac{1}{\sqrt{2}}(1+0)=\frac{1}{\sqrt{2}}[/MATH]

In fact you can easily write down the matrix representation directly from this line:



The first column is just what [MATH]\boldsymbol{v_1}[/MATH] is mapped to (in terms of the basis vectors): [MATH] \begin{pmatrix} \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} \end{pmatrix}[/MATH]
The second column is just what [MATH]\boldsymbol{v_2}[/MATH] is mapped to (in terms of the basis vectors): [MATH] \begin{pmatrix} \frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}} \end{pmatrix}[/MATH]
So the matrix for H is:

[MATH] \begin{pmatrix} \frac{1}{\sqrt{2}} &&\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} && -\frac{1}{\sqrt{2}} \end{pmatrix}[/MATH]
or

[MATH]\tfrac{1}{\sqrt{2}} \begin{pmatrix} 1 && 1\\ 1 && -1\\ \end{pmatrix}[/MATH]

 

[nasi112]

Thanks a lot lex. It was a very clear and beautiful explanation.

 

[lex]

Glad it was clear.
(I'm afraid the book was 'mystifying' rather than 'demystifying' in this case)!

 

[nasi112]

You're absolutely right. Some of the problems is very straightforward, but the book is solving them as mysterious.