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Linear Algebra with Applications by Steven J. Leon, p.257:

Theorem 5.5.2: Let $\{\textbf{u}_1, \textbf{u}_2, \ldots, \textbf{u}_n\}$ be an orthonormal basis for an inner product space $V$. If $\textbf{v} = \sum_{i=1}^{n} c_i \textbf{u}_i$, then $c_i = \langle \textbf{v}, \textbf{u}_i \rangle$.

I don't know if I need to include the proof, but it's short, so here it is:

Definition: $\delta_{ij} = \begin{cases} 1 & \text{if } i = j \\ 0 & \text{if } i \neq j \end{cases}$

Proof: $\langle \textbf{v}, \textbf{u}_i \rangle = \left< \sum_{j=1}^{n} c_j \textbf{u}_j, \textbf{u}_i \right> = \sum_{j=1}^{n} c_j \langle \textbf{u}_j, \textbf{u}_i \rangle = \sum_{j=1}^{n} c_j \delta_{ij} = c_i$

Now here's the example given in the book. He seems to be using the theorem's logic backwards. I don't get it.

Example: The vectors $ \textbf{u}_1 = \left(\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}}\right)^T \text{ and } \textbf{u}_2 = \left(\dfrac{1}{\sqrt{2}},-\dfrac{1}{\sqrt{2}}\right)^T$ form an orthonormal basis for $\mathbb{R}^2$. If $\textbf{x} \in \mathbb{R}^2$, then $\textbf{x}^T \textbf{u}_1 = \dfrac{x_1 + x_2}{\sqrt{2}} \text{ and } \textbf{x}^T \textbf{u}_2 = \dfrac{x_1 - x_2}{\sqrt{2}}$ It follows from Theorem 5.5.2 that $\textbf{x} = \dfrac{x_1 + x_2}{\sqrt{2}} \textbf{u}_1 + \dfrac{x_1 - x_2}{\sqrt{2}} \textbf{u}_2$

Isn't "$\textbf{x} = \dfrac{x_1 + x_2}{\sqrt{2}} \textbf{u}_1 + \dfrac{x_1 - x_2}{\sqrt{2}} \textbf{u}_2$" referring to "$\textbf{v} = \sum_{i=1}^{n} c_i \textbf{u}_i$", and "$\textbf{x}^T \textbf{u}_1 = \dfrac{x_1 + x_2}{\sqrt{2}} \text{ and } \textbf{x}^T \textbf{u}_2 = \dfrac{x_1 - x_2}{\sqrt{2}}$" referring to "$c_i = \langle \textbf{v}, \textbf{u}_i \rangle$"? Shouldn't the latter follow from the former? Or should the theorem state if and only if? Or am I just confused?

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    Well, if you have a theorem that says that A implies B, you can't use that theorem and B to conclude A. That's what's happening here.2012-07-04

3 Answers 3

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Every vector in your inner product space can be expressed as a unique linear combination of your basis elements.

Just think about it, when would $\langle \textbf{v}, \textbf{u}_i \rangle$ ever not equal the $i'$th coefficient?

That's why the converse is true.

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    And you get from $\textbf{v} = \sum_{i=1}^n b_i \textbf{u}_i$ to $b_i = \langle \textbf{v}, \textbf{u}_i \rangle$ by the theorem! Okay, I think I get it now. Thank you!2012-07-05
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We know that $x = c_1 u_1 + c_2 u_2$ for some $c_1, c_2$.

By the theorem, $c_1 = \left< x, u_1 \right>$ and $c_2 = \left< x, u_2 \right>$.

Thus $x = \left< x, u_1 \right> u_1 + \left< x, u_2 \right> u_2 = \frac{x_1 + x_2}{\sqrt{2}} u_1 + \frac{x_1 - x_2}{\sqrt{2}} u_2$.

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Since $\{\mathbf{u}_1, \mathbf{u}_2\}$ is a basis for $\mathbb{R}^2$, any $\mathbf{x} \in \mathbb{R}^2$ can be uniquely written as a linear combination $\mathbf{x} = a_1 \mathbf{u}_1 + a_2 \mathbf{u}_2. \tag{$\ast$}$ Taking the inner product of $(\ast)$ with $\mathbf{u}_1$, we have \begin{align*} \langle \mathbf{x}, \mathbf{u}_1 \rangle & = \langle a_1 \mathbf{u}_1 + a_2 \mathbf{u}_2, \mathbf{u}_1 \rangle \\ & = \langle a_1 \mathbf{u}_1, \mathbf{u}_1 \rangle + \langle a_2 \mathbf{u}_2, \mathbf{u}_1 \rangle \\ & = a_1 \langle \mathbf{u}_1, \mathbf{u}_1 \rangle + a_2 \langle \mathbf{u}_2, \mathbf{u}_1 \rangle \\ & = a_1 \cdot 1 + a_2 \cdot 0 \\ & = a_1. \end{align*} A similar calculation shows that $\langle \mathbf{x}, \mathbf{u}_2 \rangle = a_2.$ Therefore $a_1 = \langle \mathbf{x}, \mathbf{u}_1 \rangle = \mathbf{x}^T \mathbf{u}_1,$ $a_2 = \langle \mathbf{x}, \mathbf{u}_2 \rangle = \mathbf{x}^T \mathbf{u}_2.$

This process can be used to show that the converse of the theorem is true.

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    Yeah, that did help. Thank you!2012-07-05