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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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There is nothing strictly wrong with what he wrote; but, perhaps, it would be more clear if he instead wrote "If ${\bf x}=c_1{\bf u}_1+c_2{\bf u}_2\in\Bbb R^2$, then $$c_1={\bf x}^T{\bf u}_1 ={x_1+x_2\over\sqrt2}\ \text{and} \ c_2={\bf x}^T{\bf u}_2\ ...$$ – David Mitra Jul 4 '12 at 17:10
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. – Matt Gregory Jul 4 '12 at 17:59
up vote 2 down vote accepted

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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I can see how $\langle \textbf{v}, \textbf{u}_i \rangle$ would never not equal the $i$'th coefficient if $\textbf{u}_i$ were equal to the standard basis element $\textbf{e}_i$ because it would just be filtering out the $i$'th component of the vector. – Matt Gregory Jul 4 '12 at 19:03
I think I'm too tired and burned out to think about this properly. Why does this happen? :-P – Matt Gregory Jul 4 '12 at 19:05
Not that's not right. I'm unchecking your thing :-P What needs to be proven is that if $c_i = \langle \textbf{v}, \textbf{u}_i \rangle$, then $\textbf{v} = \sum_{j=1}^n c_i \textbf{u}_i$. – Matt Gregory Jul 4 '12 at 20:40
That follows by the fact that every $v \in V$ can be expressed as a linear combination of your basis elements. – Ink Jul 4 '12 at 21:38
Take $\textbf{v} \in V $ and suppose $c_i = \langle \textbf{v}, \textbf{u}_i \rangle$. Since $\{\textbf{u}_1,...,\textbf{u}_n\}$ is a basis for our inner product space, there exists scalars $b_1,...,b_n$ such that $\textbf{v}= \sum_{i=1}^{n} b_i \textbf{u}_i$. Hence $b_i = \langle \textbf{v}, \textbf{u}_i \rangle = c_i$. – Ink Jul 4 '12 at 22:00

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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@MattGregory: I've edited my answer to address your clarifications in the comments. Let me know if this clears up your confusion. – Henry T. Horton Jul 4 '12 at 21:14
Yeah, that did help. Thank you! – Matt Gregory Jul 5 '12 at 4:31

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