Prove Cauchy-Schwarz inequality to prove a projection onto the subspace and assume v != 0 I am asked to prove Cauchy-Schwarz inequality without using the cosine rule of the angle between u and v. 
So this is my question: 

Assume that $v\ne0$ and let $p$ be the projection of $u$ onto the subspace $V = \operatorname{span}\{v\}$. Prove that $$(u · v)^2 = \|u\|^2\|v\|^2 − \|u − p\|^2\|v\|^2.$$

Can anyone help me with this question? i look through the internet but none has specific topics on the projection onto a subspace using Cauchy-Schwarz inequality.
 A: The point is that $p=\frac{(u\cdot v)}{\|v\|^2}v$. Then 
\begin{align}
\|u-p\|^2
&=\|u\|^2+\|p\|^2-2\,u\cdot p
=\|u\|^2+\frac{(u\cdot v)^2}{\|v\|^2}-2\frac{(u\cdot v)^2}{\|v\|^2}\\ \ \\
&=\|u\|^2-\frac{(u\cdot v)^2}{\|v\|^2}.
\end{align}
So 
$$
(u\cdot v)^2=\|u\|^2\,\|v\|^2-\|u-p\|^2\,\|v\|^2
$$
A: The definition of the projection of $\mathbf u$ onto $\mathbf v$ is
$$
   \mathbf p = \frac{\mathbf u \boldsymbol{\cdot} \mathbf v}{\mathbf v \boldsymbol{\cdot} \mathbf v} \mathbf v = \frac{\mathbf u \boldsymbol{\cdot} \mathbf v}{\|\mathbf v\|^2} \mathbf v.
$$
From here, the identity you're trying to show follows from algebraic manipulation, but it can be hard to get there if you just stumble around blindly.
Begin by verifying that $\mathbf p  \boldsymbol{\cdot} \mathbf p = \mathbf p  \boldsymbol{\cdot} \mathbf u$ (getting a formula for both in terms of $\mathbf u$ and $\mathbf v$). It follows that $\mathbf p  \boldsymbol{\cdot} (\mathbf u - \mathbf p) = 0$, which gives us the Pythagorean theorem identity:
$$\|\mathbf u\|^2 = \|\mathbf p\|^2 + \|\mathbf u-\mathbf p\|^2.$$
(Just compute $\|\mathbf u\|^2 = \mathbf u \boldsymbol{\cdot}\mathbf u$, writing $\mathbf u$ as $\mathbf p + (\mathbf u - \mathbf p)$.)
The identity you want is very close to the Pythagorean theorem identity multiplied by $\|\mathbf v\|^2$, with $\|\mathbf p\|^2 = \mathbf p  \boldsymbol{\cdot} \mathbf p$ replaced by a value in terms of $\mathbf u$ and $\mathbf v$ you already know.
