Why does (ax, by) not transform like a vector under rotation?

[AMichaelson]

Homework Statement

Show that (ap1,bp2) is not a vector unless a = b.(The 1 and 2 are superscripts)
In Einstien's Gravity in a Nutshell, p 43, Zee states the above is not a vector because it doesn't transform like a vector under rotation. When I use the usual rotation matrix for rotation about the z axis (R11 = cosQ, R12 = sinQ, R21= -sinQ, R22 = cosQ), then check that length is preserved under this transformation, I get that this is in fact a vector.
Zee gives another example: (x²y, x³+y³). This, too seems to preserve length after being multiplied by the rotation matrix. What am I missing?
I would be happy to have an explanation for either example, of course.

Homework Equations

The Attempt at a Solution

I will use the first example, but make the notation easier by starting with the "vector" (non-vector?) (ax, by) = r.
r' = Rr = (axcosQ+bysinQ, -axsinQ + bycosQ)
squaring r' gets (axcosQ)² + (bysinQ)² +axbysinQcosQ + (-axsinQ)² + (bycosQ)² -axbysinQcosQ = (ax)² + (by)², which is r², so length is preserved.

Seems pretty straightforward, so: say what?[/B]

 

[TSny]

Welcome to PF!

(ax, by) = r.
r' = Rr = (axcosQ+bysinQ, -axsinQ + bycosQ)

When you write r' = Rr you are assuming that r is a vector. But you actually want to show that it is not a vector (if a ≠ b).

You are given that $(x, y)$ is a vector. So, $(x', y')$ is equal to the rotation matrix applied to $(x, y)$. That is, you know how $x$ and $y$ transform when going to the primed frame.

When you write $(ax, by)$, then $x$ and $y$ are still the components of the vector $(x, y)$. $a$ and $b$ are assumed to be scalars; so, they don't change when going to the primed frame. Therefore, the quantity $ax$ transforms as $ax$ → $ax'$. Similarly for $by$. So, you can see how $(ax, by)$ transforms. Then you can check whether or not $(ax, by)$ transforms as a vector.

 

[AMichaelson]

Thank you so much! ( Of course!)