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I am trying to understand exact linearization (feedback linearization). Through an intelligent input $u$, the output is transformed into a set of integrations:

$$v=y^{(r)},$$ where $r$ is the rth derivative of $y$

I don't see why $y$ is then linear with respect to $v$?

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    It doesn't have anything to do with the _intelligence_ of $u$. For a nonlinear system with $x$ as states, in the method of feedback linearization you find a _transformation_ such as $$z=\phi(x)$$ now if you _redefine_ the states of the system to be $z$, the output is a linear combination of $z$s and still nonlinear with respect to $x$ – 2017-01-24
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    @polfosol Thanks a lot for the comment. Could you explain a bit more. I am very new to this subject. Thanks a lot. – 2017-01-24
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    Take a look at example 1.2 on page 3 of [this lecture](http://control.ee.ethz.ch/~apnoco/Lectures2009/16-Feedback%20Linearization%20v2.pdf). Hopefully you'll get the idea – 2017-01-24
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    @polfosol Thanks a lot, – 2017-01-24

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