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For the proof of Theorem Of Pythagoras the above is given. Can anyone explain how the first step is derived? How is the length of vector u and vector v squared produce the following on the right hand side?

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    That's just $||w||^2 = w \cdot w\,$ with $w=u+v\,$. – 2017-01-08

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The dot product of two vectors is given by $$ v\cdot w = |v||w|\cos(\theta) $$ where $\theta$ is the angle between them and $|v|$ means length of $v$. When the two vectors are the same, the angle is $0$, so the formula reduces to $$ v\cdot v = |v|^2.$$

Another (equivalent) formula for the dot product is in terms of components: $$ v\cdot w = v_1w_1+v_2w_2+\ldots +v_nw_n $$

Again, when we go to the case where $v = w,$ it reduces to $$ v\cdot v = v_1^2 + v_2^2 + \ldots +v_n^2 = |v|^2.$$

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That is true by definition of the Euclidean norm.

$$\|x\| = \sqrt{x_1^2 + x_2^2 + ...+x_n^2}=\sqrt{x.x}$$

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    There are some exponents missing. – 2017-01-08
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    Oops. Thank you for pointing that out. Corrected. – 2017-01-08
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    Sure. Minor latex comment: I would use \cdot to make the dot when writing $x \cdot x$. – 2017-01-08