Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
add comment

3 Answers

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.

share|improve this answer
    
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. –  Vectk Jul 4 '12 at 21:38
1  
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 $b_i = \langle \textbf{v}, \textbf{u}_i \rangle = c_i$. –  Vectk Jul 4 '12 at 22:00
show 1 more comment

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

share|improve this answer
add comment

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.

share|improve this answer
    
@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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.