# Change of basis

1. Jul 22, 2005

### misogynisticfeminist

Say, i have a vector a, defined in a coordinate system, x-y-z and i rotate the axes by an angle theta around the z-axis, so i have my z-component invariant in this change of basis. Can someone show me why,

$$a_x' = a_x cos\theta + a_y sin\theta$$ and

$$a_x' = -a_x sin\theta + a_y cos\theta$$

i can't seem to find any place where they show how this is done.

2. Jul 22, 2005

### robphy

Draw a vector and two sets of orthogonal axes.
For each set of axes, form a right triangles with the vector as the hypotenuse, and the legs parallel to the axes.
Do a little trigononometry.
I'll post a URL with a picture, if I can find one.

Ok here's one:
http://web.umr.edu/~oci/Topic12/T12-5/cml12-5a/frame.html
(although the vector is not drawn from the origin to the little box).

(Check the subscript on the LHS of your second equation.)

Last edited: Jul 22, 2005
3. Jul 22, 2005

### LeonhardEuler

The new x-axis is a line tilted at some angle, [itex]\theta[/tex], to the horizontal. Basically, if you want [itex]a'_y[/tex], you want the distance between a point and this line. The formula for this is
$$|a'_y|= \frac {|ma_x+b-a_y|} {\sqrt{m^2+1}}$$
where the equation of the line is [itex]y=mx+b[/tex]. Since the line passes through the origin, b=0. [itex]tan(\theta)=m[/tex], so:
$$|a'_y|= \frac {|tan(\theta)a_x-a_y|} {\sqrt{tan(\theta)^2+1}}$$
$$= \frac {|tan(\theta)a_x-a_y|} {|sec(\theta)|}$$

$$= |tan(\theta)a_x-a_y|\times|cos(\theta)|$$

$$=|sin(\theta)a_x-cos(\theta)a_y|$$

Since, for [itex]\theta=0[/tex], we should have [itex]a'_y=a_y[/tex], this means the signs will only work out properly if:
$$a'_y=-sin(\theta)a_x+cos(\theta)a_y$$
The other equation is found similarly. Let me know if you want to see the derivation of this distance formula.

Last edited: Jul 22, 2005
4. Jul 22, 2005

### robphy

Here's a vector-algebraic proof.

For the first set of axes, we have unit vectors $$\hat x$$ and $$\hat y$$.
For the second set, which is rotated counterclockwise by angle $$\theta$$, we have $$\hat x'$$ and $$\hat y'$$.

Note that $$A_x= \vec A\cdot \hat x$$ and so on.

Now, compute $$A_{x'}= \vec A\cdot \hat x'$$ where one writes
$$\vec A=(\vec A \cdot \hat x) \hat x + (\vec A \cdot \hat y) \hat y=A_x\hat x + A_y\hat y$$.
So, let's plug it in:
\begin{align*} A_{x'} &= \vec A\cdot \hat x' \\ &= \left( A_x\hat x + A_y\hat y \right) \cdot \hat x' \\ &=(A_x) \hat x\cdot \hat x' + (A_y) \hat y\cdot \hat x' \\ &=(A_x) \cos\theta + (A_y) \cos(90^\circ - \theta) \\ &=(A_x) \cos\theta + (A_y) \sin\theta \\ \end{align}
Do the same for $$A_{y'}$$, and note that $$\hat x\cdot \hat y' =\cos(90^\circ+\theta)= -\sin\theta$$ and $$\hat y\cdot \hat y' =\cos\theta$$.

Last edited: Jul 22, 2005
5. Jul 22, 2005

### gerben

The original vector had an x-component and a y-component. I find it easiest to just follow those two components separately.

When the original vector gets rotated the original x-component, let’s call that X, will not be only in the direction of the x-axis anymore, the component of the rotated “original x-component“ that will still be in the direction of the x-axis is:
$$X\cos(\theta)$$
but it will now also have a component in the direction of the y-axis, which will be:
$$-X\sin(\theta)$$

The same happens with the original y component. The component of the rotated “original y component“ that will still be in the direction of the y-axis is:
$$Y\cos(\theta)$$
and the component in the direction of the x-axis will be:
$$Y\sin(\theta)$$

To get the x-component of the new vector add the parts in the direction of the x-axis:
$$X\cos(\theta) + Y\sin(\theta)$$
To get the y-component of the new vector add the parts in the direction of the y-axis:
$$-X\sin(\theta) + Y\cos(\theta)$$

6. Jul 22, 2005

### misogynisticfeminist

hey thanks alot for the help. Yup the subscript on the LHS should be y. Here's what i have done, a little algebra,

$$a_x' cos \theta = a_x$$

$$a_x' sin \theta= a_y$$

what i did was to multiply sin and cosine to both equations respectively and add them up, to get the expression for $$a_x'$$

for $$a_y'$$ i did the same thing, except that this time, comlementary angles were involved and the vector a_x points in the opposite direction.

Is that right?

edit: i'll look at the rest of the methods too. : )

Last edited: Jul 22, 2005