Prove Cauchy-Schwarz inequality to prove a projection onto the subspace and assume v != 0

Question

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.

Answer 1

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 $$ (u\cdot v)^2=\|u\|^2\,\|v\|^2-\|u-p\|^2\,\|v\|^2 $$

Answer 2

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.