Projection of vector $v$ on $u$ in terms of inner product

Question

$$ \mathrm{proj}_{v}(u) = \frac{\left \langle v,u \right \rangle}{\left \langle v,v \right \rangle}v=\left \langle \hat{v,}u \right \rangle\hat{v} $$

I am unable to follow from the second to the last inequality. Could someone kindly explain how the last equality comes about?

Answer 1

This follows from the definition of the norm as $||v||^2=\langle v,v\rangle$ and remembering that $\hat{v}=\frac{v}{||v||}$:

$$ \begin{align} \frac{\langle v,u\rangle}{\langle v,v\rangle}v&=\langle v,u\rangle\cdot\frac{v}{\langle v,v\rangle}\\ &=\langle v,u\rangle\cdot\frac{v}{||v||^2}\\ &=\left\langle \frac{v}{||v||},u\right\rangle\cdot\frac{v}{||v||}\\ &= \langle\hat{v},u\rangle\cdot\hat{v} \end{align} $$

Answer 2

$\hat{v}$ is a unit vector in the direction of $v$: $$\hat{v} = \frac{v}{|v|}; $$ Revoking that $<v,v> = |v|^2$ and property of scalar product $\alpha<a,b>=<\alpha a,b>$ we have $$\frac{<v,u>}{|v|^2}v = <\frac{v}{|v|},u>\frac{v}{|v|}=<\hat{v},u>\hat{v}. $$