Are Cobb-Douglas preferences homothetic? Our lecture defined a preference to be homothetic, if the following is true:

$$(x_1, x_2)  \thicksim (y_1, y_2) \Leftrightarrow (kx_1, kx_2) \thicksim (ky_1, ky_2)$$

Cobb-Douglas preferences can be displayed as some utility function of the following form: 

$$u(x_1, x_2) = x_1^a \cdot x_2^b$$
Therefore:
$$(x_1, x_2)  \thicksim (y_1, y_2) \\ 
\Leftrightarrow x_1^a \cdot x_2^b = y_1^a \cdot y_2^b \\
\Leftrightarrow k^ax_1^a \cdot k^bx_2^b = k^ay_1^a \cdot k^by_2^b \\
 \Leftrightarrow (kx_1, kx_2)  \thicksim (ky_1, ky_2)$$

With this argumentation the Cobb-Douglas preferences should be homothetic. 

The wikipedia article about Homothetic preferences however defined a preference to be homothetic, if they can be represented by a utility function  and the following is true:

$$ u(kx_1, kx_2) = k \cdot u(x_1, x_2)$$
And I am pretty sure, that this is not true for Cobb Douglas preferences:

$$ u(kx_1, kx_2) = (kx_1)^a (kx_2)^b = k^{a+b} x_1^a x_2^b \neq k \cdot u(x_1, x_2)$$

So what am I missing here? Are the definitions not equivalent? Did I calculate something wrong?
Note that the wikipedia article  is very specific:


  [...] defined a preference to be homothetic, if they CAN be represented by A utility function [...]


You chose a specific utility function to represent your Cobb-Douglas preferences. However there are infinitely many others. All monotonic transformations of your utility function represent the same preference. Take
$$
\hat{u}(x_1,x_2) = \left(u(x_1,x_2)\right)^{\frac{1}{a+b}} = x_1^{\frac{a}{a+b}} \cdot x_2^{\frac{b}{a+b}}.
$$
As $\hat{u}$ is a monotonic transformation of $u$, it represents the same preference. It is straightforward to check that $\hat{u}$ fullfils the condition set forth in the wiki article. So there is indeed such a utility function, that also represents the preference, hence the preference is homothetic.