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So the question is:

Let $\theta$ be a fixed real number and let
$x_1 = \begin{pmatrix} \cos\theta\\ \sin\theta \end{pmatrix}$ and $x_2 = \begin{pmatrix} -\sin\theta\\ \cos\theta \end{pmatrix}$
(a) Show that $\{x_1,x_2\}$ is an orthonormal basis for $\mathbb{R}^2$.
(b) Given a vector $y$ in $\mathbb{R}^2$, write it as a linear combination $c_1x_1+c_2x_2$.
(c) Verify that $c_1^2+c_2^2=||y||^2=y_1^2+y_2^2$

I understand how to do (a) perfectly fine. It's (b) and (c) that are giving me trouble. I'm just kind of lost, where should I begin?

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3 Answers 3

up vote 1 down vote accepted

A few (very detailed) hints. A standard result for inner products is the following:

Theorem: Let $\mathcal{B}=\left\{\mathbf{x_1},\ \cdots,\ \mathbf{x_n}\right\}$ be an orthonormal set. Then if $\mathbf{y}$ is a linear combination of the vectors of $\mathcal{B}$ we have $$\mathbf{y} = \langle\mathbf{y},\mathbf{x_1}\rangle\mathbf{x_1} + \cdots + \langle\mathbf{y},\mathbf{x_n}\rangle\mathbf{x_n}$$ Where $\langle\cdot,\cdot\rangle$ is the associated inner product.

Proof: Write $\mathbf{y}$ as a linear combination of the $\mathbf{x_i}$s. We have $$\mathbf{y} = c_1\mathbf{x_1} + \cdots + c_n\mathbf{x_n}$$ for undetermined coefficients $c_i$. Taking the inner product of both sides with each $\mathbf{x_i}$ and exploiting orthonormality, we have $$\langle\mathbf{y},\mathbf{x_i}\rangle = \langle\mathbf{x_i},\mathbf{x_i}\rangle c_i = c_i$$ The result follows. $\square$

Can you see how the method of this proof directly solves your problem? Even more simply, you can just apply the results of this theorem to question b.

For c, consider how you would calculate $\|\mathbf{y}\|^2$. You would have $$\|\mathbf{y}\|^2 = \langle\mathbf{y},\mathbf{y}\rangle = \langle c_1\mathbf{x_1}+c_2\mathbf{x_2},\ c_1\mathbf{x_1}+c_2\mathbf{x_2}\rangle$$ Using properties of inner products, how can you simplify this? Note that the method you would take to solve the above problem also generalizes to an arbitrary number of orthonormal vectors. It is the Pythagorean theorem in $n$-dimensions.

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okay I got part b, $c_1=ycos\theta +ysin\theta$ and $c_2=-ysin\theta + ycos\theta$. Now I'm still stuck on part c though. Can you give me a little more explanation on how to do it, i'm not sure i'm following your hint –  Charlie Yabben Oct 12 '12 at 2:51
    
What do we know about inner products? They are linear. Expand out the inner product step by step and see if anything simplifies. –  EuYu Oct 12 '12 at 2:53
    
so do I do ($c_1x_1+c_2x_2$)($c_1x_1+c_2x_2$)? –  Charlie Yabben Oct 12 '12 at 2:55
    
Close. It's not quite a product, but it certainly behaves just like one. I'll get you started. $$\langle c_1\mathbf{x_1}+c_2\mathbf{x_2},\ c_1\mathbf{x_1}+c_2\mathbf{x_2}\rangle = c_1\langle \mathbf{x_1},\ c_1\mathbf{x_1}+c_2\mathbf{x_2}\rangle + c_2\langle \mathbf{x_2},\ c_1\mathbf{x_1}+c_2\mathbf{x_2}\rangle$$ Can you see how to continue? –  EuYu Oct 12 '12 at 2:57
    
Oh do you just multiply it out and combine like terms and such? –  Charlie Yabben Oct 12 '12 at 3:01

For b, write $y = c_1 x_1 + c_2 x_2$, equate the two components of the vector equation (remembering that the $x_i$ are vectors and the $c_i$ are reals), and solve for $c_1$ and $c_2$. Once you have these, you can verify part c.

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what exactly do you mean by "equate the two components". I'm not following on that part. –  Charlie Yabben Oct 12 '12 at 2:46

For part (b): Suppose $y=(y_1,y_2)=c_1x_1+c_2x_2$. This sets up the equation

$ \begin{pmatrix} y_1 \\ y_2 \end{pmatrix} = \begin{pmatrix} a & -b \\ b & a \end{pmatrix} \times \begin{pmatrix} c_1 \\ c_2 \end{pmatrix} $

where $a=\cos \theta, b = \sin \theta$. Can you solve for $c_1,c_2$ above,perhaps by inverting the $2\times 2$ matrix?

Part (c) should follow immediately, because the $2\times 2$ matrix is a rotation matrix. So it leaves $y$ unchanged in length.

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Thanks @C.Y for the edit! –  Ganesh Oct 12 '12 at 2:51

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