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We are given the following basis: $$\{{b}_{0}, {b}_{1}\}=\{\begin{pmatrix}1\\ 1\\ \end{pmatrix}, \begin{pmatrix}1\\ -1\\ \end{pmatrix}\}$$ Normalized with ${\ell}^{2}$ norm: $${\stackrel{~}{b}}_{0}=\frac{1}{\sqrt{2}}\begin{pmatrix}1\\ 1\\ \end{pmatrix}$$ $${\stackrel{~}{b}}_{1}=\frac{1}{\sqrt{2}}\begin{pmatrix}1\\ -1\\ \end{pmatrix}$$ Normalized with ${\ell}^{1}$ norm: $${\stackrel{~}{b}}_{0}=\frac{1}{2}\begin{pmatrix}1\\ 1\\ \end{pmatrix}$$ $${\stackrel{~}{b}}_{1}=\frac{1}{2}\begin{pmatrix}1\\ -1\\ \end{pmatrix}$$
Standard basis for ${\mathbb{R}}^{2}$ , also referred to as ${\ell}^{2}(\left[0 , 1\right])$ : $${b}_{0}=\begin{pmatrix}1\\ 0\\ \end{pmatrix}$$ $${b}_{1}=\begin{pmatrix}0\\ 1\\ \end{pmatrix}$$ $${b}_{0}\cdot {b}_{1}=\sum_{i=0}^{1} {b}_{0}(i){b}_{1}(i)=1\times 0+0\times 1=0$$
Now we have the following basis and relationship: $$\{\begin{pmatrix}1\\ 1\\ \end{pmatrix}, \begin{pmatrix}1\\ -1\\ \end{pmatrix}\}=\{{h}_{0}, {h}_{1}\}$$ $${h}_{0}\cdot {h}_{1}=1\times 1+1\times -1=0$$
Pulling the previous two sections (definitions) together, we arrive at the most important and useful basis type:
$$\{{b}_{0}, {b}_{2}\}=\{\begin{pmatrix}1\\ 0\\ \end{pmatrix}, \begin{pmatrix}0\\ 1\\ \end{pmatrix}\}$$
$$\{{b}_{0}, {b}_{2}\}=\{\begin{pmatrix}1\\ 1\\ \end{pmatrix}, \begin{pmatrix}1\\ -1\\ \end{pmatrix}\}$$
$$\{{b}_{0}, {b}_{2}\}=\{\frac{1}{\sqrt{2}}\begin{pmatrix}1\\ 1\\ \end{pmatrix}, \frac{1}{\sqrt{2}}\begin{pmatrix}1\\ -1\\ \end{pmatrix}\}$$
Orthonormal bases are very easy to deal with! If $\{{b}_{i}\}$ is an orthonormal basis, we can write for any $x$
Given the following basis: $$\{{b}_{0}, {b}_{1}\}=\{\frac{1}{\sqrt{2}}\begin{pmatrix}1\\ 1\\ \end{pmatrix}, \frac{1}{\sqrt{2}}\begin{pmatrix}1\\ -1\\ \end{pmatrix}\}$$ represent $x=\begin{pmatrix}3\\ 2\\ \end{pmatrix}$
We are given the basis
$$(n, )$$∞
Let $\{{b}_{i}\}$ be an orthonormal basis for a Hilbert space $H$ . Then, for any $x\in H$ we can write
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